Question

8A : Consider the titration, 10 mL of NaCN(Ka=4.9X10^-10) with .380M HCl. Calculate pH at 0.0mL HCl, 11.82mL HCl.

8B How many mL HCl is needed to reach the equivalence point and what is the pH at this point?

Answer 1

First of all, I think the concentration of the NaCN solution is missing here, because we can not know the pH of this solution without knowing the amount of NaCN dissolved in 10 ml of solvent. If you are using a titration to know the concentration of an unknown solution of NaCN, then you will need the HCl volume reached in the equivalence point to calculate the amount of NaCN in the original solution (and this is the question 8B).

Having said that, I will assume that our NaCN solution is 0.760M, for the purpose of illustrate the most relevant cases during a titration

NaCN + 0.0 mL HCl

The NaCN is a salt. It dissociates and you will have its conjugated base in solution, CN^-. Cyanide will dissociate in water, producing a basic solution, according to:

CN^- + H₃O⁺ ?HCN + H₂O

CN^- + H₂O ? HCN + OH^-                           Kb = Kw/Ka

Ka = 4.9 x 10^-10; Kb = 2.04 x 10^-4

	CN- +	H2O	?	HCN +	OH-
Initial	0.760M	-		-	-
Change	-X	-		+X	+X
Equilibrium	0.760-X			+X	+X

Kb=(x)(x)/ (0.760-x)                      Since X is less than 5% from 0.380 we can simplify the eq.

0.760*Kb=X²                                     Solving for X

X=1.25*10^-2

pOH = -log (1.25*10^-2) = 1.90

pH = 14- pOH = 12.10

NaCN + 11.82 mL HCl

We calculate the number of moles that are consumed in this reaction to know if the equivalence point has been reached or not

	CN- +	H3O+	?	HCN +	H2O
Initial (n)	0.0076	-		-	-
Change (n)		-0.0044
Reaction (n)	0.0032	-		+0.0044	+0.0044
Concentration (n/V)	0.146M	-		0.201M

The concentrations are dictated by the new volume= 10ml NaCN + 11,82mL HCl

Since we have still CN^- remaining in the solution, an equilibrium is established

	CN- +	H2O	?	HCN +	OH-
Initial (n)	0.146M	-		0.201M	-
Change (n)	-X			+X	+X
Equilibrium	0.146-X			0.201+X	X

Kb=(0,201+x)(x)/ (0,146-x)                         Since X is less than 5% from M we can simplify the eq.

0.146*Kb/0.201=X                                         Solving for X

X=1.48*10^-4

pOH = -log (1.48*10^-4) = 3.83

pH = 14- pOH = 10.17

Equivalence point.

In the equivalence point we know that the number of moles of CN- are neutralized by an equal number of moles of H3O+, provided by HCL. We know that our solution is 0.760M, and in 10 mL we have 0.0076 n that reacts with 0.0076 n of HCl to produce HCN. we calculate the volume from this amount:

[HCl]= 0,380 n/1l

0.0076n *1l/ 0.380 n = 20 mL HCl.

Now our solution contains only HCN, because all the CN- was consumed by HCl. A new equilibrium is established.

We have formed 0.0076 moles of HCN that are dissolved in 30mL of water. the new concentration is 0.253M

	HCN +	H2O	?	CN- +	H3O+
Initial	0.253M	-		-	-
Change (n)	-X			+X	+X
Equilibrium	0.253-x	-		X	X

Ka=(x)(x)/ (0.253-x)                       Since X is less than 5% from 0.380 we can simplify the eq.

0.253*Ka=X²                                     Solving for X

X=1.11*10^-5

pH = -log (1.11*10^-5) = 4.95