a cell containing a silver indicator electrode and a reference electrode with a constant potential of $0\mathrm{.}175V.$ The reference

electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal.

The solubility constant of TIBr is $K_{sp}=3\mathrm{.}6{10}^{-6}$ and the solubility constant of AgBr is $K_{sp}=5\mathrm{.}0{10}^{-13}.$

Which precipitate forms first?

TIBr

AgBr

Which of the expressions shows how the cell potential, $E,$ depends on $A{g}^{+}?$

$E=[0\mathrm{.}799(V)-0\mathrm{.}05916log([A{g}^{+}])]-0\mathrm{.}175V$

$E=0\mathrm{.}175V-[0\mathrm{.}799(V)-0\mathrm{.}05916log([1],\left[\begin{array}{c}A{g}^{+}\end{array}\right])]$

$E=0\mathrm{.}175V-[0\mathrm{.}799(V)-0\mathrm{.}05916log([A{g}^{+}])]$

$E=[0\mathrm{.}799(V)-0\mathrm{.}05916log(\frac{1}{[A{g}^{+}]})]-0\mathrm{.}175V$

Calculate the first and second equivalence points of the titration.

first equivalence point: $mL$

second equivalence point: $mL$

What is the cell potential after each of the given volumes of $0\mathrm{.}200$MNaBr have been added?

$1\mathrm{.}0mL,E=$

$26\mathrm{.}5mL,E=$

$26\mathrm{.}5mL.E=$

$27\mathrm{.}4mL,E=$

$27\mathrm{.}8mL,E=$

$45\mathrm{.}8mL,E=$

$55\mathrm{.}0mL,E=$

$64\mathrm{.}0mL,E=$

A solution prepared by mixing $55\mathrm{.}0mL$ of $0\mathrm{.}100MAgN{O}_3$ and $55\mathrm{.}0mL$ of $0\mathrm{.}100MTIN{O}_3$ was titrated with $0\mathrm{.}200$MNaBr ir a cell containing a silver indicator electrode and a reference electrode with a constant potential of $0\mathrm{.}175V.$ The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of T$1$Br is $K_{sp}=3\mathrm{.}6{10}^{-6}$ and the solubility constant of AgBr is $K_{sp}=5\mathrm{.}0{10}^{-13}.$

Which precipitate forms first?

TIBr

AgBr

Which of the expressions shows how the cell potential, $E,$ depends on $A{g}^{+}?$

$E=[0\mathrm{.}799(V)-0\mathrm{.}05916log([A{g}^{+}])]-0\mathrm{.}175V$

$E=0\mathrm{.}175V-[0\mathrm{.}799(V)-0\mathrm{.}05916log(\frac{1}{[A{g}^{+}]})]$

$E=0\mathrm{.}175V-[0\mathrm{.}799(V)-0\mathrm{.}05916log([A{g}^{+}])]$

$E=[0\mathrm{.}799(V)-0\mathrm{.}05916log(\frac{1}{[A{g}^{+}]})]-0\mathrm{.}175V$

Calculate the first and second equivalence points of the titration.

$\frac{}{\frac{1}{2}}$

first equivalence point: $mL$

second equivalence point:

$mL$

What is the cell potential after each of the given volumes of $0\mathrm{.}200$MNaBr have been added?