A buffer that contains 0.123 M acid, HY, and 0.211 M of its conjugate base, Y−, has a pH of 8.67. What is the pH after 0.0018 mol of Ba(OH)2 is added to 0.325 L of this solution?

pH after adding 0.0018 mol Ba(OH)₂ = 8.69

Explanation:

For this problem, 1st determine Ka for acid HY. This can be determined using the Henderson Hasselbalch Equation or by applying the Ka expresson ratio against the hydronium ion concentration.

Given pH (before adding Ba(OH)₂) = 8.67 => [H⁺] = 10⁻⁸°⁶⁷ = 2.138 x 10⁻⁹M

Ka(HY) = [H⁺][Y⁻]/[HY] = (2.138 x 10⁻⁹)(0.211)/(0.123) = 3.67 x 10⁻⁹

or, using the Henderson-Hasselbalch Equation...

pH = pKa + log([Base]/[Acid]) => pKa = pH - log([Base]/[Acid])

pKa = 8.69 - log([0.211]/[0.123]) = 8.43 => Ka = 10⁻⁸°⁴³ = 3.67 x 10⁻⁹

When 0.0018 mole Ba(OH)₂ is added into the buffer solution, 2(0.0018 mol) OH⁻ is delivered into solution = 0.0036 mol OH⁻. The OH⁻ will react with the available H⁺ forming H₂O essentially removing it from the acid equilibrium causing the acid equilibrium to shift right to deliver more H⁺. This will continue until all of the 0.0036 mole of OH⁻ has been consumed. Then a new acid equilibrium will be established with new concentrations of HY, H⁺ & Y⁻.  Using the ICE table solution approach will best illustrate this effect.

           HY      =>     H⁺      +     Y⁻

I =>   0.123M        10⁻⁸°⁶⁷       0.211M <= Original Equilibrium Conc

C=> -0.0036M        |      +0.0036M <=Right Shift=>Dec HY+Inc Y⁻

E=>  0.1194M        [H⁺]           0.2146M <= New Equilibrium Conc

Ka = [H⁺][Y⁻]/[HY] => [H⁺] = Ka·[HY]/[Y⁻] = (3.67 x 10⁻⁹)(0.1194)/(0.2146) = 2.04 x 10⁻⁹M => pH(new) = -log[H⁺] = -log(2.04 x 10⁻⁹) = 8.69 (new pH)*.

*Note: In general, for buffer solutions, the pH increases when OH⁻ is added and decreases if H⁺ is added. In this case, addition of OH⁻ does result in an increase in pH as expected.

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