Chemistry homework help! Desperate!

Ive been trying this problem for hours now. If someone could please help me in any way, I would greatly appreciate it! PLEASE :-)


A buffer was made by mixing 0.1464 moles of HCN with 0.1459 moles of NaCN and diluting to exactly 1 liter. What will be the pH after addition of 10.00 mL of 0.3444 M RbOH to 50.00 mL of the buffer? Ka(HCN) = 4.900e-10. Note: only a portion of the original buffer is used in the second part of the problem.

The answer is 9.752, but I dont know HOW to get it. Please. Help.

RbOH + HCN ---> RbCN + H2O

So, the HCN amount goes down and the CN- amount goes up. Let's get moles of RbOH:

(0.3444 mol/L) (0.010 L) = 0.003444 mol

Now, get moles of HCN and CN-

HCN ---> (0.1464 mol/L) (0.050 L) = 0.00732 mol
CN- ---> (0.1459 mol /L) (0.050 L) = 0.007295 mol

Let's get moles after reacting with the hydroxide:

HCN ---> 0.00732 mol minus 0.003444 mol = 0.003876 mol
CN- ---> 0.007295 mol plus 0.003444 mol = 0.010739 mol

Now, we need the pKa of HCN:

pKa = -log 4.900e-10 = 9.3098

Now, fire up the Henderson-Hasselbalch equation:

pH = pKa + log [salt] / [acid]

pH = 9.3098 + log (0.010739 / 0.003876)

pH = 9.3098 + 0.44258 = 9.75238 (which rounds off to 9.752)

Notice that I did not bother to get molarities for the log portion of the H-H equation. The two values are both in 60 mL, so that part just cancels out.

Good problem.