# Calculate the pH of a buffer that contains 1.00 M hydrocyanic acid (HCN) and 0.60 M sodium cyanide (NaCN).?

a. What is the pH after 10 mL of 10.0 M NaOH is added to 1 L of this buffer.

### 1 Answer

- hcbiochemLv 74 months ago
My text gives a Ka for HCN = 6.17x10^-10, so pKa = 9.21

HCN <--> H+ + CN-

There are a couple of ways to deal with the first part of this. One is to write the expression for Ka, plug in the concentrations that you have and solve for [H+]. Then calculate the pH from that.

Ka = [H+][CN-]/[HCN] = 6.17X10^-10

The other is to use the Henderson-Hasselbalch equation:

pH = pKa + log [CN-]/[HCN]

pH = 9.21 + log (0.60 / 1.00) = 8.99

a. The initial buffer contains 1.00 mole of HCN and 0.60 mol of CN-. The NaOH added contains 0.010 L X 10.0 mol/L = 0.10 mol NaOH. The added OH- quantitatively neutralizes 0.10 mol of the HCN forming an additional 0.10 mol of CN-. So, after the addition, the solution contains 0.90 mol HCN and 0.70 mol CN-.

pH = 9.21 + log (0.70/0.90) = 9.10

Note: I used moles of HCN and CN- directly rather than dividing each by the total volume to calculate [HCN] and [CN-]. The ratio of moles of each buffer component is the same as the ratio of molarities of each component.