Question

calculate the ph of a solution made by mixing 20.0 ml of 0.200 m h2so3, 10.0 ml of 0.120m naoh, 10.0ml of 0.150 m hcl, and 10.0ml of 0.200 m na2so3. (pka1: 1.857, pka2: 7.172 for h2so3)

Answer 1

There are several reactions:

H₂SO₃+H₂O=HSO₃^-+H₃⁺O (pka1=1.857, ka1=0.0139) ka1=10^-pka1. [H⁺]=0.00053 (1)

HSO₃^-+H₂O=SO₃^2-+H₃⁺O (pka2=7.172, ka1=6.72x10^-8) ka2=10^-pka2. [H⁺?]=0.000000597 (2)

NaOH+H2O=Na++OH-+H₂O (complete dissociation) (mol NaOH=0.0012)

HCl+H₂O=H₃⁺O+Cl^- (completely dissociation) (mol HCl=0.0015)

NaOH+HCl=NaCl+H2O ( all of NaOH is consumed by this reaction. This salt is neutral pH=7 since it proceeds from both strong acid and base). (3)

According the information we have 0.0012 mol NaOH and 0.0015 mol HCl. For this reaction it remains 0.0003 mol HCl.

On the oher hand, we have a solution of basic salt (Na2SO3). So, we have to calculated the H-concentration of this salt. The reactions are:

Na₂SO₃=2Na⁺+SO₃^2- .

2Na⁺+H2O=NaOH+2H⁺ (because of NaOH is a strong-base, we can neglect this reaction).

SO₃^2-+H₂O=HSO₃^-+OH^- (kb=kw/ka2=1.48x10^-7) (4)

HSO₃^-+H₂O=SO₃^2-+OH^- (kb=kw/ka1=7.19x10^-13) (5)

From (1), (2), (3)

mol H⁺=0.00053+0.0003+0.000000597 mol =0.00083.

From (4), (5)

mol OH^-=0.000000172 mol (these are consumed with iqual quantity of H⁺ to produce water).

At the end we have [H⁺]=(0.00083)/(0.05 L) (because we have to plus all of volume) = 0.0166 M H⁺.

ph=-log([H])=1.78.