Answer:
pH = 7.233
Explanation:
Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.
NaHSO₃ reacts with NaOH thus:
NaHSO₃ + NaOH → Na₂SO₃ + H₂O
50.0 mL of 1.00 M NaOH are:
0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced 0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:
NaHSO₃: 0.208 mol - 0.050 mol = <em>0.158 mol</em>
Na₂SO₃: 0.134 mol + 0.050 mol = <em>0.184 mol</em>
<em> </em>
As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:
pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]
pH = 7.167 + log₁₀ [0.184] / [0.158]
<em>pH = 7.233</em>
