20. You make 100.0 mL of a 0.44M Acetic Acid-Acetate buffer (pKa = 4.74) at a pH of 4.4. You then add 2.0 mL of a 0.60M solution of Sr(OH)2 to the buffer. What is the new pH? 0.601 Sr(OH) = 0.002K 4.4 = 4.74 + -0.34 = log ( log | bo base acid base acid ( base, acid 0.457 - base= 0.457 acid acid + 0.457acid = 0.44 M 1.457 acid = 0.44 acid= 0.302 M x 0.1L = 0.0302 mol base: 0.138 Mx x 0.1 L =0,0138 mul X = 0.0012 mol = 4.74 +logl pH PH= 4.45 10.0138 +0.0012 0,0302-0.002