Question

A chemist titrates 160.0 mL of a 0.3065 M cyanic acid (HCNO) solution with 0.4994 M NaOH solution at 25 °C. Calculate the pH at equivalence. The pK of cyanic acid is 3.46 Round your answer to 2 decimal places.

Answer 1

Answer: The pH at equivalence for the given solution is 8.37.

Step-by-step explanation:

We know that,

`pK_(a) = -log K_(a)`

3.46 =
`-log K_(a)`

`K_(a) = 3.467 * 10^(-4)`

Now, we will calculate the volume of NaOH required to reach the equivalence point as follows.

`M_((HCNO)) * V_((HCNO)) = M_((NaOH)) * V_((NaOH))`

`0.3065 M * 160.0 mL = 0.4994M * V_((NaOH))`

`V_((NaOH))` = 98.1978 mL

The given data is as follows.

M(HCNO) = 0.3065 M

V(HCNO) = 160 mL

M(NaOH) = 0.4994 M

V(NaOH) = 98.1978 mL

So, moles of HCNO will be calculated as follows.

mol(HCNO) =
`M(HCNO) * V(HCNO)`

mol(HCNO) =
`0.3065 M * 160 mL`

= 49.04 mmol

Now, the moles of NAOh will be calculated as follows.

mol(NaOH) =
`M(NaOH) * V(NaOH)`

mol(NaOH) =
`0.4994 M * 98.1978 mL`

= 49.04 mmol

This means that, 49.04 mmol of both HCNO and NaOH will react to form
`CNO^(-)` and
`H_(2)O`.

Here,
`CNO^(-)` is strong base . So,

`CNO^(-)` formed = 49.04 mmol

Total volume of the solution is as follows.

Volume of Solution = 160 + 98.1978 = 258.1978 mL

And,

`K_(b)` of
`CNO^(-)` =
`(K_(w))/(K_(a))`

=
`(1 * 10^(-14))/(3.467 * 10^(-4))`

=
`2.884 * 10^(-11)`

The concentration of
`CNO^(-)` is as follows.

c =
`(49.04 mmol)/(258.1978 mL)`

= 0.1899M

Also,

`CNO^(-) + H_(2)O \rightarrow HCNO + OH^(-)`

Initial: 0.1899 0 0

Equilibm:0.1899 - x x x

`K_(b) = ([HCNO][OH^(-)])/([CNO^(-)])`

`K_(b) = (x * x)/((c - x))`

We assume that x can be ignored as compared to c . Hence, above formula can be rewritten as follows.

`K_(b) = (x * x)/(c)`

so, x =
`\sqrt (K_(b) * c)`

x =
`\sqrt ((2.884 * 10^(-11)) * 0.1899)`

=
`2.34 * 10^(-6)`

Here, c is much greater than x, this means that our assumption is correct .

so, x =
`2.34 * 10^(-6)` M

`[OH^(-)] = x = 2.34 * 10^(-6)` M

As,

pOH =
`-log [OH^(-)]`

=
`-log (2.34 * 10^(-6))`

= 5.6307

Also, pH = 14 - pOH

= 14 - 5.6307

= 8.3693

or, = 8.37 (approx)

Thus, we can conclude that the pH at equivalence for the given solution is 8.37.