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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.

User JJaun
by
7.7k points

1 Answer

4 votes

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.

User Naki
by
8.3k points
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