Question

wwN1 1934: HWK 15 G calculator-Google Search .com/sw/mod/ .php?id=688900&isStudentz1 estion 17 of 20 (1 point) x Incorrect x Incorrect COAST Tutorial Problem How much 5.90 M NaOH must be added to 650.0 mL of a buffer that is 0.0205 Macetic acid and 0.0280 M sodium acetate to raise the pH to 5.757 Number 8.71 mL When you mix an acid, such as acetic acid with its conjugate base (acetate), you create a buffer. When you have a buffer, you can use the Henderson-Hasselbach equation, but you need to look up the pKa first C Ty Again 16 17 1 18920 Activity Grade: 20 Points Possible
How much 5.90 M NaOH must be added to 650.0 mL of a buffer that is 0.0205 M acetic acid and 0.0280 M sodium acetate to raise pH to 5.75?

Answer 1

Let volume of NaOH be V mL

mol of NaOH added = 5.90*V mmol

Before adding NaOH

Before Reaction:

mol of CH3COO- = 0.0280 M *650.0 mL

mol of CH3COO- = 18.20 mmol

mol of CH3COOH = 0.0205 M *650.0 mL

mol of CH3COOH = 13.325 mmol

5.9*V NaOH will react with 5.9*V of CH3COOH to form extra 5.9*V of base

After adding NaOH

mol of CH3COOH = 13.325-5.9*V mmol

mol of CH3COO- = 18.2+5.9*V mmol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.7447

use:

pH = pKa + log {[conjugate base]/[acid]}

5.75 = 4.7447+log {[CH3COO-]/[CH3COOH]}

log {[CH3COO-]/[CH3COOH]} = 1.0053

[CH3COO-]/[CH3COOH] = 10.1221

So,

(18.2+5.9*V)/(13.325-5.9*V) = 10.1221

18.2+5.9*V = 134.8776 - 59.7206*V

(5.9+59.7206)*V = 134.8776-18.2

V = 1.778 mL

Answer: 1.778 mL