Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN (Ka of HCN = 6.2 x 10-10) with a) 4.20 mL of 0.438 M HClO4 and b) 11.82 mL of 0.438 M HClO4. What is the pH at the equivalence point with 0.438 M HClO4?

The data:

millimoles NaCN = 50.00*0.100 = 5.00
mmoles HClO4 = 4.20*0.438 = 1.8396 (I know that's too many significant figures but you can always drop the extra ones at the end.)
mmoles HClO4 = 11.82*0.438 = 5.177
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I'm not copying all of the zeros; they sometimes mess up the spacing on this ICE chart.
.........NaCN + HClO4 ==> HCN + NaClO4
initial.. 5......0.........0.......0
add...........1.8396................
change..-1.8396..-1.8396...1.8396..etc
equil....3.16...0..........1.8396...etc

You can see for part a that you have a weak acid (HCN) and a salt of the weak acid in solution (NaCN) so this is a buffered solution. Use the Henderson-Hasselbalch equation and solve for pH.

part b. Set up an ICE chart as in part a to see what you have. I think you will have 0.177 mmoles HClO4 remaining in soln along with 5 mmoles HCN. So that is a weak acid plus a common ion (H^+) from the HClO4. You make another ICE chart with HCN==> H^+ + CN^-
substitute into Ka (remember to add the H^+ common ion), and solve for pH.

The last part for pH at equivalence point, that is determined by the pH of the NaCN salt. Set up the hydrolysis and use an ICE chart for that. You will need to determine the (CN^-) at the equivalence point which I will call c.
........CN^- + HOH ==> HCN = OH^-
.........c...............0.....0
change...-x.............x.......x
equil....c-x............x.......x

Kb = (Kw/Ka) = (HCN)(OH^-)/(CN^-)
Kw = 1E-14
Ka = Ka for HCN
(HCN) + (OH^-) = x
(CN^-) = c-x
Solve for x = OH^- and convert to pH.
Post your work if you get stuck.

I converted every thing to mols instead of milli mols so for the first part got

pH=9.208 + log(1-(4.20/11.42)/(4.20/11.42)

so my pH= 9.44 when 4.20 mL of 0.438 M HClO4 is added?

That looks good to me.

Great I think I can figure the rest out! Thank you for your help!

To calculate the pH of a solution, we need to consider the reactions and equilibrium that occur when different substances are mixed together. In this case, we need to calculate the pH after mixing the given solutions. Let's break it down step by step:

a) Mixing 50.00 mL of 0.100 M NaCN with 4.20 mL of 0.438 M HClO4:
First, we need to determine if any reaction occurs between NaCN and HClO4. NaCN is a salt of a weak acid (HCN), while HClO4 is a strong acid. Therefore, the reaction between them is not significant, and we can consider NaCN as a spectator ion.

Now, let's calculate the amount of H+ ions added to the solution by 4.20 mL of 0.438 M HClO4 (assuming complete dissociation):
moles of H+ = (volume x concentration) = (0.00420 L x 0.438 mol/L) = 0.0018372 mol

Since NaCN doesn't contribute any H+ ions, but it produces OH- ions, we need to calculate the amount of OH- ions:
moles of OH- = 0.1 mol/L x 0.05 L = 0.005 mol

Next, we need to compare the amount of H+ and OH- ions to determine if the resulting solution is acidic or basic:
[H+] = (0.0018372 - 0.005) mol = 0.0018322 mol
[OH-] = 0.005 mol

The equilibrium reaction of water (H2O) can help us calculate the pH:
H2O ⇌ H+ + OH-

Given that the concentration of H2O remains much larger than both H+ and OH- concentrations, we neglect the small changes in [H+] and [OH-] due to the ionization of water.

Using the equation Kw = [H+][OH-], we can determine the concentration of H+ ions:
Kw = [H+][OH-]
1.0 x 10^-14 = (0.0018322)(0.005)
[H+] = (1.0 x 10^-14)/(0.0018322) = 5.460 x 10^-12 mol/L

Finally, we can find the pH using the equation pH = -log[H+]:
pH = -log(5.460 x 10^-12) = 11.262

Therefore, the pH of the solution after mixing 50.00 mL of 0.100 M NaCN with 4.20 mL of 0.438 M HClO4 is approximately 11.262.

b) Mixing 50.00 mL of 0.100 M NaCN with 11.82 mL of 0.438 M HClO4:
Following a similar process as above, we can calculate the resulting pH. However, since the volume of HClO4 has increased, the amount of added H+ ions will be different.

moles of H+ = (0.01182 L x 0.438 mol/L) = 0.00517236 mol

Using the same approach as before, we determine the [H+] and [OH-] concentrations:
[H+] = (0.00517236 - 0.005) mol = 0.00017236 mol
[OH-] = 0.005 mol

Kw = [H+][OH-]
1.0 x 10^-14 = (0.00017236)(0.005)
[H+] = (1.0 x 10^-14)/(0.00017236) = 5.802 x 10^-10 mol/L

pH = -log(5.802 x 10^-10) = 9.237

Therefore, the pH of the solution after mixing 50.00 mL of 0.100 M NaCN with 11.82 mL of 0.438 M HClO4 is approximately 9.237.

c) The pH at the equivalence point with 0.438 M HClO4:
At the equivalence point, the moles of H+ ions added will react completely with the moles of OH- ions to form water. Therefore, the resulting solution will be neutral, and the pH will be 7.

To summarize:
a) The pH is approximately 11.262 after mixing 50.00 mL of 0.100 M NaCN with 4.20 mL of 0.438 M HClO4.
b) The pH is approximately 9.237 after mixing 50.00 mL of 0.100 M NaCN with 11.82 mL of 0.438 M HClO4.
c) The pH at the equivalence point with 0.438 M HClO4 is 7.