Please help me in the steps to this problem?

Question

The pH of a 1.00 x 10^-3 M solution of HOCN is 2.77 at 25*C. Calculate Ka from this result.

What are the steps involved?

We're solving for the equilibrium constant not the rate constant... it's a good way to start

According to my textbook in the solutions manual it states the Ka as being 3.5 x 10^-4 so I have to assume there is a way to solve this..

Answer 1

There are two problems with this question. The most serious is that the problem can't be answered as posed. To see this, let's assume that the Ka of HOCN is infinite (i.e., HOCN completely dissociates in water to form H+ and OCN- ions, with no HOCN remaining). In that case, a 10^-3M solution of HOCN would yield a solution of H+ and OCN- ions, each with concentration 10^-3 M.

Now, pH is defined as -log[H+], where [H+] means the concentration of H+ ions. (Strictly speaking, pH is defined in terms of the chemical activity of H+ ions, but this is usually glossed over in introductory chem classes.) Our 10^-3M solution of H+ ions would therefore have a pH of 3. In order to have a pH of 2.77, one would need a H+ concentration of 10^-2.77M, but there aren't enough H+ ions in a 10^-3M solution of HOCN to provide this many! It is therefore impossible to solve this question.

The second problem is that cyanic acid (HOCN) is unstable in water. It reacts with water to form carbon dioxide and ammonia via the reaction:

HOCN + H2O -> CO2 + NH3.

CO2 and NH3 will, in turn, react with water and the ultimate pH will be a result of the equilibria among the resulting species:

CO2 + H2O -> H2CO3
H2CO3 -> H+ + HCO3-
HCO3- -> H+ _ CO3--

NH3 + H2O -> NH4+ + OH-

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Added to respond to questioner's second comment.

Answer keys have been known to be wrong, and that's the case in this instance.

Let's set up the problem as though it can be solved, though.

The dissociation reaction for this compound can be written as:

HOCN <-> (H+) + (OCN-)

The dissociation constant is given by:

Ka = [H+]*[OCN-]/[HOCN]

Initially (i.e., before any dissociation), the concentration of HOCN was 10^-3M. Every H+ ion that is formed reduces the number of HOCN molecules by one, so at equilibrium:

[HOCN] = 10^-3M - [H+]

This is simply a statement of mass balance (conservation of mass).

We also know from the dissociation reaction (or from charge balance considerations, i.e., the solution must be electrically neutral) that [H+] = [OCN-]

We can use these two constraints to eliminate [OCN-] and [HOCN] from the expression for Ka:

Ka = [H+]^2/{10^-3 - [H+]}

You are told that the pH is equal to 2.77. From the definition of pH, we have that:

-log([H+]) = 2.77

[H+] = 10^(-2.77) = 1.698 * 10^-3

Substitute this into the expression for Ka and simplify:

Ka = {1.698 * 10^-3}^2/{10^-3 - 1.698 * 10^-3}

Ka = {2.884 * 10^-6}/{-6.98 * 10^-4}

Ka = -4.132 * 10^-3

But dissociation constants MUST be positive numbers, so something is wrong with the data you have been given.

We can calculate what the pH of a 10^-3 M solution of a monoprotic acid with a Ka of 3.5 * 10^-4 (the value given in your answer key) should be. Start again with the expression derived above for Ka as a function of [H+]:

Ka = [H+]^2/{10^-3 - [H+]} = 3.5 * 10^-4

[H+]^2 = 3.5*10^-4 * {10^-3 - [H+]}

[H+]^2 + 3.5*10^-4 * [H+] - 3.5*10^-7 = 0

This is a quadratic equation in the unknown [H+]. Using the quadratic formula to solve for [H+] yields the roots:

[H+] = -7.919 * 10^-4 or +4.419 * 10^-4. The negative root has no physical meaning; the positive root is the one we want

[H+] = 4.419*10^-4 M

pH = -log(4.419*10^-4) = 3.355.

If HOCN were actually stable in water, and if it's Ka were 3.5*10^-4, then a 1M solution of HOCN would have a pH of 3.355.

Answer 2

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Answer 3

sorry guy i forgot these. it is a good question. i will wait to see the responses

Answer 4

H2O2 + 2C- + N+ -> I2 + 2H2O
rate = d[H2O2]/dt = k[H2O2]m[I-]n
I2 + 2S2O3 2- -> 2I- + S4O6 2-

DONT ASK ME IF IT RIGHT