3.) Hydrocyanic acid, HCN, is a weak acid. The following is the equilibrium equa

Question

3.)

Hydrocyanic acid, HCN, is a weak acid. The following is the equilibrium equation for its reaction with water:

HCN(aq) + H2O(l) <----------> H3O+ + CN-(aq)     Ka = 6.2 x 10-10

What is the pH of a 1.87 M HCN solution?

10.)

What is the hydronium ion concentration, [H3O+] in a 2.60 M NH4Cl solution? Consider the following information to help you answer this question:

Ammonium chloride is a water-soluble salt. Like all water-soluble ionic compounds, it dissociates completely when dissolved in water:

                 H2O
NH4Cl(s) ----------> NH4+(aq) + Cl-(aq)

Since the Cl- ion is the conjugate base of a strong acid (HCl), it is too weak of a base to react with water:

Cl-(aq) + H2O(l) ----------> N.R.

Note: N.R. means “No Reaction”.

However, the NH4+ ion is the conjugate acid of a weak base (NH3). Therefore, it is a strong enough acid to react with water:

NH4+(aq) + H2O(l) <----------> H3O+(aq) + NH3(aq)

The above equilibrium reaction is the one for which you need to set up and solve an ICE table. However, the Ka for NH4+ is normally not published in tables. It doesn’t’ need to be, because you can get it from two other values that are published: the Kb for NH3, and KW for the ionization of water.

NH3(aq) + H2O(l) <----------> NH4+(aq) + OH-(aq)      Kb = 1.8 x 10-5

2H2O(l) <----------> H3O+(aq) + OH-(aq)     KW = 1.0 x 10-14

By combining the above two equations in the appropriate way, you can produce the reaction for which the equilibrium constant is desired. Once you have the required equilibrium constant, you can set up the ICE table in the usual way. Note that the initial concentration of NH4+ is the same as the concentration given for NH4Cl. This is because the dissociation reaction proceeds to 100% completion, and the reaction has one-to-one stoichiometry. For every mole of NH4Cl, you get a mole of NH4+ (and a mole of Cl-).

Explanation / Answer

Given: The dissociation of HCN would be as follows:

HCN(aq) + H2O(l) <----------> H3O+ + CN-(aq), Ka = 6.2 x 10-10, [HCN] = 1.87 M

To find: pH of HCN

Solution: We should set up an ICE table to obtain [H3O+] and pH.

HCN(aq) + H2O(l) <----------> H3O+ + CN-(aq)

I 1.87 M - 0 0

C -x   +x      +x

E 1.87-x x x

Ka of HCN = 6.2 x 10-10

[H3O+][CN-]/[HCN] = 6.2 x 10-10

x.x/(1.87-x) = 6.2 x 10-10

Since HCN is a weak acid, x in the denominator can be neglected.

x2/(1.87) = 6.2 x 10-10

x2 = 1.87 x 6.2 x 10-10

x2 = 11.59 x 10-10

x= sqrt(11.59 x 10-10)

x = 3.40 x 10-5

[H3O+] = 3.40 x 10-5

pH = -log[H3O+] = -log(3.40 x 10-5)

pH = 5 -log(3.40) = 4.47

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