Question

500 mL of a buffer solution contains 0.050 mol NaHSO3 and 0.031 mol Na2SO3.

(a) What is the pH of the solution?

(b) Write the net ionic equation for the reaction that occurs when NaOH is added to this buffer.

(c) Calculate the new pH after 10. mL of 1.0 M NaOH is added to the buffer solution.

d) calculate the buffer capacity

(f) Calculate the new pH after 10. mL of 1.0 M NaOH is added to 500. mL of pure water.

Answer 1

a)

NaHSO3 --> Na+ + HSO3-

Na2SO3 --< 2Na+ + SO3-2

then

pH = pKa + log(SO3-2 / HSO3-)

pKa for second ionizaiton of H2SO3

pKa = 6.91

pH = 6.91+ log(0.031 / 0.05) = 6.702

pH = 6.702

b)

net ionic equations:

acid + base: salt + water

H+ + OH- --> H2O(l)

c)

new pH after M = 1 of NaOH and V = 10 mL

mmol of NAOH = MV = 1*10 = 10 mmol of OH- added = 10*10^-3 = 0.01 mol of OH-

HSO3- new = 0.05 - 0.01 = 0.04

SO3-2 new = 0.031 + 0.01 = 0.041

then

pH = 6.91+ log(0.041/0.04) = 6.92

d)

buffer capacity --> ph must change at least 1 unit

so:

7.91 = 6.91 + log(ratio)

ratio = 10 mL

f)

if we a add 10 mL of 1 M

then

VT = 500+10 = 510 mL = 0.51 L

mol of base = MV = 10*1 = 10 mmol = 0.01 mol of OH-

M = mol/V = 0.01/510 = 0.0000196078

pOH = -log(OH-) = -log(0.0000196078 = 4.70757

pH = !4-4.70757

pH = 9.29243