If you have a 1.0 L buffer containing 0.208 M NaHSO3 and 0.134 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH?

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 = 0.158 mol

Na₂SO₃: 0.134 mol + 0.050 mol = 0.184 mol

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]

pH = 7.233

where's the diagram?