I need help with this link Experiment 11 Buffer pH and Property (Work Sheet) docx Experiment 11 Buffer pH and Property (Worksheet) Student name (Print) I Suggestion of Use (1) This is a dry lab (2) Instructors may assign different problems for different students in the same group to work on in the lab (3) If preferred, instructors may randomize those numbers in the given problems and then assign to students II Objectives (1) Identify necessary components for the constitution of a buffer (2) Calculate buffer pH using HendersonHasselbalch equation (3) Calculate buffer pH after the addition of strong acid base III Principles, Procedures, and Practices Buffers are widely used in chemical reaction where solution pH is required to be constant With the addition of limited amount of acid base, buffer solution pH does not change significantly III 1 Constitution of a buffer To constitute a buffer, a pair of conjugate weak acid and weak base is required The conjugate weak acid and the conjugate weak base has a difference of only one H For example, HC 2 H 3 O (acetic acid) and C 2 H 3 O (acetate) will be t constitute a buffer Your instructor will select a problem for you to work on (Different problems may be assigned to different students in the same group ) Question Identify which of the following combination will be able to constitute a buffer Problem 1 Problem 2 Problem 3 HCN (aq) HSCN (aq) yes no HNO 3 (aq) NO 3 (aq) yes no HClO 3 (aq) ClO 3 (aq) yes no HC 2 H 3 O (aq) C 2 H 3 O (aq) yes no HNO 2 (aq) NaNO 2 (aq) yes no HClO (aq) KClO (aq) yes no Cl (aq) HCl (aq) yes no NH 4 (aq) NH 3 (aq) yes no NaH 2 PO 4 (aq) Na 2 HPO 4 (aq) yes no H 2 S (aq) Na 2 S (aq) yes no H 2 SO 4 (aq) Na 2 SO 4 (aq) yes no H 2 CO 3 (aq) NaHCO 3 (aq) yes no III 2 Buffer solution pH calculation For buffer solution pH related calculation, HendersonHasselbalch equation is often used pH p K a log base ini acid ini However, one must always keep in mind that HendersonHasselbalch equation is good for use only when, 0 1 base ini acid ini 10 Example What is the pH of a buffer solution composed of 0 12 M NaNO 2 and 0 15 M HNO 2 at 25C It is known K a for HNO 2 is 4 510 4 at 25C Solution HNO 2 is a weak acid The NO 2 derived from NaNO 2 is a weak base HNO 2 and NO 2 are a conjugate pair base ini NO 2 ini 0 12M acid ini HNO 2 ini 0 15M pH p K a log base ini acid ini log 4 5 10 4 log 0 12 0 15 3 35 log0 80 3 35 ( 0 10) 3 25 Attention Sometimes, mass may be given instead of mole or molarity In that case, you have to calculate molarity first Your instructor will select a problem for you to work on (Different problems may be assigned to different students in the same group ) Question Calculate the buffer solution pH Problem 1 Problem 2 Problem 3 What is the pH of a buffer solution composed of 0 0200 M boric acid and 0 020 M sodium borate at 25 C Boric acid K a 5 810 10 A 500 ml buffer solution contains 15 0 g NH 4 Cl and 6 15 g 30 (by mass) NH 3 solution at 25 C What is the pH of this buffer NH 4 K a 5 810 10 MM NH3 17 03g mol, MM NH4Cl 53 49g mol What is the pH of a 250 ml buffer containing 0 15 mol NaH 2 PO 4 and 0 22 mol Na 2 HPO 4 at 25 C K a for H 2 PO 4 is 6 210 8 Solution III 3 Buffer solution Preparation for a Required pH Buffer preparation has two common ways (1) direct mixing conjugate acid and base and (2) partial neutralization Way 1 Direct mixing conjugate acid and base Way 2 Partial neutralization Example It is known nitrous acid (HNO 2 ) pKa 3 35 If you intend to use HNO 2 and NaNO 2 to prepare a buffer solution at pH 4 00, what is molar ration of NaNO 2 and HNO 2 should be To a solution containing 15 0 g HNO 2 , what mass of NaNO 2 you need to add Solution pH p K a log base ini acid ini log base ini acid ini pH p K a 4 00 3 35 0 65 Hence, base ini acid ini 10 0 65 4 7 NaN O 2 ini HNO 2 ini 4 7 mass of NaNO 2 mass of HNO 2 NaN O 2 ini HNO 2 ini MM of NaNO 2 MM of HNO 2 4 7 MM of NaNO 2 MM of HNO 2 4 7 69 00 47 01 6 9 mass of NaNO 2 mass of NaNO 2 mass of HNO 2 mass of HNO 2 6 915 0g 1 0 10 2 g Example It is known nitrous acid (HNO 2 ) pKa 3 35 If you have 200 ml 0 100 M HNO 2 solution, what volume of 0 500 M NaOH you need to add in order to have a buffer with pH at 3 00 Solution HNO 2 aq NaOH aq NaNO 2 aq H 2 O(l) With partial neutralization, the solution will include HNO 2 and NO 2 , which can constitute a buffer pH p K a log base ini acid ini log base ini acid ini pH p K a 3 00 3 35 0 35 Hence, base ini acid ini 10 0 35 0 45 NaN O 2 ini HNO 2 ini mol of NaN O 2 mole of HNO 2 4 7 Assuming x ml of 0 100 M NaOH is required to add HNO 2 aq NaOH aq NaNO 2 aq H 2 O l Based on above chemical equation, moles of NaOH quals to moles of NaNO 2 generated 0 500M x 1000 L 0 100M 200 1000 L 0 500M x 1000 L 4 7 5 00 10 4 xmol 2 00 10 2 mol 5 00 10 4 xmol 4 7 x 33 0 ml Your instructor will select a problem for you to work on (Different problems may be assigned to different students in the same group ) Question Problem 1 Problem 2 Problem 3 It is known H 2 PO 4 pKa 7 20 If you intend to use NaH 2 PO 4 and Na 2 HPO 4 to make a buffer with a pH at 7 00, to a solution containing 10 0 g NaH 2 PO 4 , what mass of Na 2 HPO 4 should be added MM NaH2PO4 120 0 g mol, MM Na2HPO4 142 0 g mol It is known H 2 PO 4 pKa 7 20 If you have 150 ml 0 150 M NaH 2 PO 4 solution, what volume of 0 200 M NaOH you need to add to make a buffer with pH at 8 00 NaH 2 PO 4 (aq) NaOH(aq) Na 2 HPO 4 (aq) H 2 O(l) It is known H 2 PO 4 pKa 7 20 If you intend to use NaH 2 PO 4 and Na 2 HPO 4 to make a buffer with a pH at 7 50, to a solution containing 22 0 g NaH 2 PO 4 , what mass of Na 2 HPO 4 should be added MM NaH2PO4 120 0 g mol, MM Na2HPO4 142 0 g mol Solution III 4 Buffer property Buffer solution can react with both acid and base to keep pH relative stable, while water does not have this property Adding strong acid to pure water Example Pure water pH 7 00 at 25 C If 1 00 ml 0 100 M HCl is added into 100 ml pure water, what is the pH Solution HCl is a strong acid It 100 ionize after being added into water HCl moles 1 00 1000 L0 100M 1 00 10 4 mol H HCl moles 100 1000 L 1 00 10 4 mol 0 100L 1 00 10 3 M pH log 1 00 10 3 3 00 Adding strong base to pure water Example Pure water pH 7 00 at 25 C If 1 00 ml 0 100 M NaOH is added into 100 ml pure water, what is the pH Solution NaOH is a strong acid It 100 ionize after being added into water NaOH moles 1 00 1000 L0 100M 1 00 10 4 mol OH NaOH moles 100 1000 L 1 00 10 4 mol 0 100L 1 00 10 3 M pOH log 1 00 10 3 3 00 pH 14 00 pOH 14 00 3 00 11 00 Adding strong acid into buffer Example A 100 ml buffer solution containing 0 100 M HNO 2 and 0 100 M NaNO 2 What is the pH before and after adding 1 00 ml 0 100 M HCl HNO 2 pKa 3 35 Solution Before adding HCl pH p K a log base ini acid ini 3 35 log 0 100M 0 100M 3 35 After adding HCl HCl moles 1 00 1000 L0 100M 1 00 10 4 mol NO 2 aq HCl aq HNO 2 aq Cl (aq) Reaction between NO 2 and HCl is 1 1 molar ratio reaction Moles of HCl reacted equal to moles of NO 2 consumed and moles of HNO 2 generated NO 2 original moles 100 1000 L0 100M 1 00 10 2 mol NO 2 remaining moles NO 2 original moles HCl moles 1 00 10 2 mol 1 00 10 4 mol 9 9 10 3 mol HNO 2 original moles 100 1000 L0 100M 1 00 10 2 mol HNO 2 remaining moles HNO 2 original moles HCl moles 1 00 10 2 mol 1 00 10 4 mol 1 01 10 2 mol pH p K a log base ini acid ini 3 35 log 9 9 10 3 mol 1 01 10 2 mol 3 35 0 0087 3 34 Adding strong base into buffer Example A 100 ml buffer solution containing 0 100 M HNO 2 and 0 100 M NaNO 2 What is the pH after adding 1 00 ml 0 100 M NaOH HNO 2 pKa 3 35 Solution NaOH moles 1 00 1000 L0 100M 1 00 10 4 mol HNO 2 aq NaOH aq NaNO 2 aq H 2 O(aq) Reaction between HNO 2 and NaOH is 1 1 molar ratio reaction Moles of NaOH reacted equal to moles of HNO 2 consumed and moles of NO 2 generated HNO 2 original moles 100 1000 L0 100M 1 00 10 2 mol HNO 2 remaining moles HNO 2 original moles NaOH moles 1 00 10 2 mol 1 00 10 4 mol 9 9 10 3 mol NO 2 original moles 100 1000 L0 100M 1 00 10 2 mol NO 2 remaining moles NO 2 original moles NaOH moles 1 00 10 2 mol 1 00 10 4 mol 1 0 10 2 mol pH p K a log base ini acid ini 3 35 log 1 01 10 2 mol 9 9 10 3 mol 3 35 0 0087 3 36 Summary Adding 1 00 ml 0 100 M HCl into 100ml water results a pH change from 7 00 to 3 00 (DpH 4 00) Adding 1 00 ml 0 100 M NaOH into 100ml water results a pH change from 7 00 to 11 00 (DpH 4 00) Adding 1 00 ml 0 100 M HCl into 100ml buffer results a pH change from 3 35 to 3 34 (DpH 0 01) Adding 1 00 ml 0 100 M NaOH into 100ml water results a pH change from 3 35 to 3 36 (DpH 0 01) Attention If too much acid base is added resulting in base ini acid ini 0 1 or 10, one cant use HendersonHasselbalch equation to calculate buffer pH any more Instead, common ion effect with I C E table should be used for pH calculation Your instructor will select a problem for you to work on (Different problems may be assigned to different students in the same group ) Questions Problem 1 Problem 2 Problem 3 A 100 ml buffer solution containing 0 200 M HC 2 H 3 O 2 (acetic acid) and 0 300 M NaC 2 H 3 O 2 What is the pH before and after adding 2 00 ml 0 100 M HCl Acetic acid pK a 4 74 A 100 ml buffer solution containing 0 200 M HC 2 H 3 O 2 (acetic acid) and 0 150 M NaC 2 H 3 O 2 What is the pH before and after adding 3 00 ml 0 200 M HCl Acetic acid pK a 4 74 A 100 ml buffer solution containing 0 250 M HC 2 H 3 O 2 (acetic acid) and 0 200 M NaC 2 H 3 O 2 What is the pH before and after adding 5 00 ml 0 200 M NaOH Acetic acid pK a 4 74 Solution

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I need help with this link Experiment 11. Buffer pH and Property (Work Sheet).docx Experiment 11. Buffer pH and Property (Worksheet) Student name (Print):_______________________ I.

I need help with this link

Experiment 11. Buffer pH and Property (Work Sheet).docx

Experiment 11. Buffer pH and Property (Worksheet)

Student name (Print):_______________________
I. Suggestion of Use (1) This is a dry lab. (2) Instructors may assign different problems for different students in the same group to work on in the lab. (3) If preferred, instructors may randomize those numbers in the given problems and then assign to students.

II. Objectives (1) Identify necessary components for the constitution of a buffer; (2) Calculate buffer pH using HendersonHasselbalch equation; (3) Calculate buffer pH after the addition of strong acid/base.

III. Principles, Procedures, and Practices
Buffers are widely used in chemical reaction where solution pH is required to be constant. With the addition of limited amount of acid/base, buffer solution pH does not change significantly. III.1 Constitution of a buffer To constitute a buffer, a pair of conjugate weak acid and weak base is required. The conjugate weak acid and the conjugate weak base has a difference of only one H⁺. For example, HC₂H₃O (acetic acid) and C₂H₃O^- (acetate) will be t constitute a buffer.
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Question: Identify which of the following combination will be able to constitute a buffer.
Problem 1		Problem 2				Problem 3
HCN (aq) + HSCN (aq)	yes; no.	HNO₃ (aq) + NO₃^- (aq)		yes; no.		HClO₃ (aq) + ClO₃^- (aq)	yes; no.
HC₂H₃O (aq) + C₂H₃O^- (aq)	yes; no.	HNO₂ (aq) +NaNO₂(aq)		yes; no.		HClO (aq) + KClO (aq)	yes; no.
Cl^- (aq) + HCl (aq)	yes; no.	NH₄⁺(aq) + NH₃ (aq)		yes; no.		NaH₂PO₄ (aq) + Na₂HPO₄ (aq)	yes; no.
H₂S (aq) + Na₂S (aq)	yes; no.	H₂SO₄ (aq) +Na₂SO₄(aq)		yes; no.		H₂CO₃ (aq) +NaHCO₃(aq)	yes; no.

III.2 Buffer solution pH calculation For buffer solution pH-related calculation, HendersonHasselbalch equation is often used. pH=pKa+log[base]ini[acid]ini However, one must always keep in mind that HendersonHasselbalch equation is good for use only when, 0.1<[base]ini[acid]ini<10 Example: What is the pH of a buffer solution composed of 0.12 M NaNO₂ and 0.15 M HNO₂ at 25C? It is known K_a for HNO₂ is 4.510^-4 at 25C. Solution: HNO₂ is a weak acid. The NO₂^- derived from NaNO₂ is a weak base. HNO₂ and NO₂^- are a conjugate pair. [base]_ini=[NO₂^-]_ini=0.12M; [acid]_ini=[HNO₂]_ini=0.15M pH=pKa+log[base]ini[acid]ini=-log4.510-4+log0.120.15 =3.35+log0.80 = 3.35+(-0.10)=3.25 Attention: Sometimes, mass may be given instead of mole or molarity. In that case, you have to calculate molarity first.
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Question: Calculate the buffer solution pH.
Problem 1		Problem 2				Problem 3
What is the pH of a buffer solution composed of 0.0200 M boric acid and 0.020 M sodium borate at 25 C? Boric acid K_a=5.810^-10.		A 500-ml buffer solution contains 15.0 g NH₄Cl and 6.15 g 30% (by mass) NH₃ solution at 25 C. What is the pH of this buffer? NH₄⁺ K_a=5.810^-10. MM_NH3=17.03g/mol, MM_NH4Cl=53.49g/mol.				What is the pH of a 250-ml buffer containing 0.15 mol NaH₂PO₄ and 0.22 mol Na₂HPO₄ at 25 C. K_a for H₂PO₄^- is 6.210^-8.
Solution:
III.3 Buffer solution Preparation for a Required pH Buffer preparation has two common ways: (1) direct mixing conjugate acid and base; and (2) partial neutralization.
Way 1 Direct mixing conjugate acid and base				Way 2- Partial neutralization
Example: It is known nitrous acid (HNO₂) pKa = 3.35. If you intend to use HNO₂ and NaNO₂ to prepare a buffer solution at pH = 4.00, what is molar ration of NaNO₂ and HNO₂ should be? To a solution containing 15.0 g HNO₂, what mass of NaNO₂ you need to add? Solution: pH=pKa+log[base]ini[acid]ini log[base]ini[acid]ini=pH-pKa=4.00-3.35=0.65 Hence, [base]ini[acid]ini=100.65=4.7 [NaNO2]ini[HNO2]ini=4.7 mass of NaNO2mass of HNO2=NaNO2iniHNO2iniMM of NaNO2MM of HNO2 =4.7MM of NaNO2MM of HNO2=4.769.0047.01 = 6.9 mass of NaNO2=mass of NaNO2mass of HNO2mass of HNO2 =6.915.0g=1.0102g				Example: It is known nitrous acid (HNO₂) pKa = 3.35. If you have 200 ml 0.100 M HNO₂ solution, what volume of 0.500 M NaOH you need to add in order to have a buffer with pH at 3.00? Solution: HNO2aq+NaOHaqNaNO2aq+H2O(l) With partial neutralization, the solution will include HNO₂ and NO₂^-, which can constitute a buffer. pH=pKa+log[base]ini[acid]ini log[base]ini[acid]ini=pH-pKa=3.00-3.35=-0.35 Hence, [base]ini[acid]ini=10-0.35=0.45 [NaNO2]ini[HNO2]ini=mol of NaNO2mole of HNO2=4.7 Assuming x ml of 0.100 M NaOH is required to add. HNO2aq+NaOHaqNaNO2aq+H2Ol Based on above chemical equation, moles of NaOH quals to moles of NaNO₂ generated. 0.500Mx1000L0.100M2001000L-0.500Mx1000L=4.7 5.0010-4xmol2.0010-2mol-5.0010-4xmol=4.7 x=33.0 ml
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Question:
Problem 1			Problem 2		Problem 3
It is known H₂PO₄^- pKa=7.20. If you intend to use NaH₂PO₄ and Na₂HPO₄ to make a buffer with a pH at 7.00, to a solution containing 10.0 g NaH₂PO₄, what mass of Na₂HPO₄ should be added? MM_NaH2PO4=120.0 g/mol, MM_Na2HPO4=142.0 g/mol.			It is known H₂PO₄^- pKa=7.20. If you have 150 ml 0.150 M NaH₂PO₄ solution, what volume of 0.200 M NaOH you need to add to make a buffer with pH at 8.00? NaH₂PO₄(aq) + NaOH(aq) Na₂HPO₄(aq) + H₂O(l)		It is known H₂PO₄^- pKa=7.20. If you intend to use NaH₂PO₄ and Na₂HPO₄ to make a buffer with a pH at 7.50, to a solution containing 22.0 g NaH₂PO₄, what mass of Na₂HPO₄ should be added? MM_NaH2PO4=120.0 g/mol, MM_Na2HPO4=142.0 g/mol.
Solution:
III.4 Buffer property Buffer solution can react with both acid and base to keep pH relative stable, while water does not have this property.
Adding strong acid to pure water Example: Pure water pH=7.00 at 25 C. If 1.00 ml 0.100 M HCl is added into 100 ml pure water, what is the pH? Solution: HCl is a strong acid. It 100% ionize after being added into water. HCl moles=1.001000L0.100M=1.0010-4mol H+=HCl moles1001000L=1.0010-4mol0.100L=1.0010-3M pH=-log1.0010-3=3.00
Adding strong base to pure water Example: Pure water pH=7.00 at 25 C. If 1.00 ml 0.100 M NaOH is added into 100 ml pure water, what is the pH? Solution: NaOH is a strong acid. It 100% ionize after being added into water. NaOH moles=1.001000L0.100M=1.0010-4mol OH-=NaOH moles1001000L=1.0010-4mol0.100L=1.0010-3M pOH=-log1.0010-3=3.00 pH=14.00-pOH=14.00-3.00=11.00
Adding strong acid into buffer Example: A 100-ml buffer solution containing 0.100 M HNO₂ and 0.100 M NaNO₂. What is the pH before and after adding 1.00 ml 0.100 M HCl? HNO₂ pKa = 3.35. Solution: Before adding HCl pH=pKa+log[base]ini[acid]ini=3.35+log0.100M 0.100M =3.35 After adding HCl HCl moles=1.001000L0.100M=1.0010-4mol NO2-aq+HClaqHNO2aq+Cl-(aq) Reaction between NO₂^- and HCl is 1:1 molar ratio reaction. Moles of HCl reacted equal to moles of NO₂^- consumed and moles of HNO₂ generated. NO2- original moles=1001000L0.100M=1.0010-2mol NO2- remaining moles=NO2- original moles-HCl moles =1.0010-2mol-1.0010-4mol=9.910-3mol HNO2 original moles=1001000L0.100M=1.0010-2mol HNO2 remaining moles=HNO2 original moles+HCl moles =1.0010-2mol+1.0010-4mol=1.0110-2mol pH=pKa+log[base]ini[acid]ini=3.35+log9.910-3mol 1.0110-2mol =3.35-0.0087=3.34
Adding strong base into buffer Example: A 100-ml buffer solution containing 0.100 M HNO₂ and 0.100 M NaNO₂. What is the pH after adding 1.00 ml 0.100 M NaOH? HNO₂ pKa = 3.35. Solution: NaOH moles=1.001000L0.100M=1.0010-4mol HNO2aq+NaOHaqNaNO2aq+H2O(aq) Reaction between HNO₂ and NaOH is 1:1 molar ratio reaction. Moles of NaOH reacted equal to moles of HNO₂ consumed and moles of NO₂^- generated. HNO2 original moles=1001000L0.100M=1.0010-2mol HNO2 remaining moles=HNO2 original moles-NaOH moles =1.0010-2mol-1.0010-4mol=9.910-3mol NO2- original moles=1001000L0.100M=1.0010-2mol NO2- remaining moles=NO2- original moles+NaOH moles =1.0010-2mol+1.0010-4mol=1.010-2mol pH=pKa+log[base]ini[acid]ini=3.35+log1.0110-2mol9.910-3mol =3.35+0.0087=3.36
Summary Adding 1.00 ml 0.100 M HCl into 100ml water results a pH change from 7.00 to 3.00. (DpH = -4.00) Adding 1.00 ml 0.100 M NaOH into 100ml water results a pH change from 7.00 to 11.00. (DpH = +4.00) Adding 1.00 ml 0.100 M HCl into 100ml buffer results a pH change from 3.35 to 3.34. (DpH = -0.01) Adding 1.00 ml 0.100 M NaOH into 100ml water results a pH change from 3.35 to 3.36. (DpH = +0.01) Attention: If too much acid/base is added resulting in [base]_ini/[acid]_ini <0.1 or >10, one cant use HendersonHasselbalch equation to calculate buffer pH any more. Instead, common ion effect with I.C.E. table should be used for pH calculation.
Your instructor will select a problem for you to work on. (Different problems may be assigned to different students in the same group.)
Questions:
Problem 1			Problem 2		Problem 3
A 100-ml buffer solution containing 0.200 M HC₂H₃O₂ (acetic acid) and 0.300 M NaC₂H₃O₂. What is the pH before and after adding 2.00 ml 0.100 M HCl? Acetic acid pK_a = 4.74.			A 100-ml buffer solution containing 0.200 M HC₂H₃O₂ (acetic acid) and 0.150 M NaC₂H₃O₂. What is the pH before and after adding 3.00 ml 0.200 M HCl? Acetic acid pK_a = 4.74.		A 100-ml buffer solution containing 0.250 M HC₂H₃O₂ (acetic acid) and 0.200 M NaC₂H₃O₂. What is the pH before and after adding 5.00 ml 0.200 M NaOH? Acetic acid pK_a = 4.74.
Solution: