# How to calculate the pH of the neutralisation of HCN with excess of KOH?

Write the reaction between $\pu{0.05 mol}$ $\ce{HCN (aq)}$ and $\pu{0.08 mol}$ $\ce{KOH (aq)}$ with $V = \pu{500 mL}$ and then calculate the pH.

So I wrote the reaction: $$\ce{HCN (aq) + KOH (aq) <=> KCN (aq) + H2O}$$

Begin: \begin{align} n_\mathrm{begin}(\ce{HCN}) &= \pu{0.05 mol}\\ n_\mathrm{begin}(\ce{KOH}) &= \pu{0.08 mol}\\ n_\mathrm{begin}(\ce{KCN}) &= -\\ n_\mathrm{begin}(\ce{H2O}) &= -\\ \end{align}

End: \begin{align} n_\mathrm{end}(\ce{HCN}) &= -\\ n_\mathrm{end}(\ce{KOH}) &= \pu{0.08 mol} - \pu{0.05 mol} = \pu{0.03 mol}\\ n_\mathrm{end}(\ce{KCN}) &= \pu{0.05 mol}\\ n_\mathrm{end}(\ce{H2O}) &= \pu{0.05 mol}\\ \end{align}

The product of the reaction is a salt of a strong base ($\ce{CN-}$) completely dissociated in water: $$\ce{KCN (aq) -> K+ (aq) + CN- (aq)}$$

And $$\ce{CN- (aq) + H2O -> HCN (aq) + OH- (aq)}$$

Knowing this I calculated the $$[\ce{KCN}] = \frac{\pu{0.05 mol}}{\pu{0.5 L}} = \pu{0.1 M};$$

Then I used the equation: $$[\ce{OH-}]_1 = \sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}} \cdot [\ce{KCN}]} = \pu{1.27x10^-3 M}$$

Since I still have some $\ce{KOH}$, it dissociates completely creating $\ce{OH-}$ ions, so: \begin{align} [\ce{OH-}]_2 &= [\ce{KOH}]\\ [\ce{OH-}]_2 &= \frac{\pu{0.03 mol}}{\pu{0.5 L}} = \pu{0.06 M} \end{align}

Finally \begin{align} [\ce{OH-}] &= [\ce{OH-}]_1 + [\ce{OH-}]_2 \\ &= \pu{1.27x10^-3 M} + \pu{0.06 M} = \pu{0.06127 M}\\ \ce{pOH} &= -\log[\ce{OH-}] = -\log(0.0617) = 1.213\\ \ce{pH} &= 14 - \ce{pOH} = 12.787\\ \end{align}

Is it right to calculate the $[\ce{OH-}]$ in this way?

I based my calcualtions on this equation:
Knowing that the amounts of substance in moles are defined by $Z = Z_1 + Z_2$ in $K$ liters of solution, we can consider this concentration to be $$[X] = \frac{Z}{K} = \frac{Z_1 + Z_2}{K} = \frac{Z_1}{K} + \frac{Z_2}{K} = [X_1] + [X_2]$$

While you've generally done a good job, you've made an important error. The hydrolysis of the cyanide ion should be written as an equilibrium reaction:

$$\ce{CN-(aq) + H2O(l) <=> HCN(aq) + OH-(aq)}$$

That's because $\ce{HCN}$ is a very weak acid and its conjugated base a weak base (homework: look up the $K_b$ of $\ce{CN-}$).

Now, as you duly noted, there's excess $\ce{KOH}$ in the solution, which dissociates completely. The $\ce{OH-}$ resulting from that dissociation now push the equilibrium:

$$K_b=\frac{[\ce{HCN}]\times[\ce{OH-}]}{[\ce{CN-}]}$$

to the left. As a result, the contribution of the $[\ce{OH-}]$ coming from the hydrolysis of the cyanide is negligible and can be ignored. Only the $[\ce{OH-}]$ from the excess $\ce{KOH}$ is to be counted. So $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$.

We can confirm that numerically by assuming $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$ and with:

$$\frac{[\ce{HCN}]}{[\ce{CN-}]}=\frac{K_b}{[\ce{OH-}]}\approx 0.0027$$

So neglibibly little cyanide is present as the acid $\ce{HCN}$.

As regards the matter of adding molarities, in some instances that is allowed.

Say we add $n_1$ moles of $X$ to $V\ \mathrm{L}$, that would give a molarity $M_1=\frac{n_1}{V}$. At a later stage we add $n_2$ moles of $X$, that molarity would be $M_2=\frac{n_2}{V}$.

The total molarity would be:

$$M=M_1+M_2=\frac{n_1}{V}+\frac{n_2}{V}=\frac{n_1+n_2}{V}$$

But if we were to mix volumes of solutions it would be:

$$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$$

Now they are no longer simply additive.

• Since CN- is the conjugate base of a weak acid shouldn't be a strong base? Nov 30, 2017 at 16:57
• The $pK_b=4.79$, so quite weak. Thing e.g. acetic acid/acetate.
– Gert
Nov 30, 2017 at 17:00
• So considering it a weak base the concentration of OH- is defined only by the KOH concentration, due to the fact that CN- has not enough force to "steal" an H from the water. Right? Nov 30, 2017 at 17:12
• That's a poetic way of putting it but basically correct. Better to just look at the equilibrium constant expression, that tells you $[\ce{HCN}]\approx 0$, if excess hydroxide is present. Without excess $\ce{KOH}$ things would be different. The excess hydroxide suppresses the cyanide hydrolysation.
– Gert
Nov 30, 2017 at 17:16
• That's the point I was looking for. On the internet and even on my university book there are only examples where no reagent remain at the end of the reaction. If I can ask you, is the cyanide hydrolysation suppressed because of the stronger Kb of the OH- base? Nov 30, 2017 at 18:00

The OH− resulting from KOH (0.06 M) that dissociation now push the equilibrium:

Kb=[HCN]×[OH−]/[CN−]

The contribution of the [OH−] coming from the hydrolysis of the cyanide can be ignored. Only the [OH−] from the excess KOH is to be counted. So [OH−]≈0.06 mol/L.

From hydrolise of CN-, we have [HCN]=[OH−], so we have:

Kb=[HCN][OH−]/[CN−]=[OH−][OH−](from KOH)/[CN−]=[OH−]x0.1 M /0.06 M [OH−]≈0.000027

Finally [OH−]=[OH−]+[OH−]=0.06 M + 0.000027 M =0.060027 M pH=14-(−log[OH−])=14−(-log0.060027)=12.778

• Useful links for text and formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial. Jun 18 at 9:29
• Apart from the formatting improvements to be made, this does not really answer the question, it only repeats the calculation. It also contradicts the accepted answer. Concentrations can generally only be added under very certain conditions. How much sense measures a number of 0.000027 in this context anyway. Jun 18 at 23:29

First of all your calculation is based on the assumption that HCN is a strong acid, which it's not. Check for ionisation constant of HCN. And yes you can add moles directly because moles are discrete particles, even if they are in solution. But you can't add concentration or molarity because it's volume dependent and when you consider volumes, you'll eventually end up adding moles again.

• I know that HCN is a weak base (its Ka is 6.2x10^(-3)) but I was thinking that, since it reacts with a strong base, it would be completly neutralized. Am I wrong? Nov 30, 2017 at 17:00
• In infinite time, maybe yes. But when you calculate pH with a pH meter, it gives you the instantaneous value. And if I'm not wrong, there's a buffer formation as well Nov 30, 2017 at 17:07
• What do you mean in 'infinite time'? Do you realise how fast ionic reactions in water really are? And no, there is no buffer effect here: that would only be the case with partial neutralisation, so there's significant $CN^-$ and $HCN$ present.
– Gert
Nov 30, 2017 at 17:13