# Calculating pH of weak base and strong acid: Why is K(a) not required?

Question: 35mL of 1.5M $$\ce{HCN}$$, a weak acid(K(a)= $$6.2*10^{-10}$$) is mixed with 25mL of 2.5M $$\ce{KOH}$$. Calculate the pH of final solution.

The solution given in my textbook is as follows:-

1. $$\ce{KOH}$$ completely dissociates to give 0.062 mol $$\ce{OH-}$$(The $$\ce{K+}$$ becomes irrelevant)

2. 0.052 mol of $$\ce{HCN}$$

3.The equation is $$\ce{HCN + OH- ⇌ CN- + H2O}$$

1. So 0.062-0.052 mol=0.01 mol of $$\ce{OH-}$$ is left over

2. so 0.01 mol/(0.035l+0.025l)=0.17M of $$\ce{OH-}$$

6.pOH=0.76 so pH=13.24

7.Also, $$\ce{CN-}$$ is a weak base relative to $$\ce{OH-}$$ so we consider it negligible in the calculation.

1. Finally, it says that K(a) was not required and it was just to distract.

However, I think K(a) is actually required because $$\ce{HCN + OH- ⇌ CN- + H2O}$$ is a reversible reaction and I don't understand how all of the 0.052 mol of $$\ce{HCN}$$ can react with $$\ce{OH-}$$(and in turn, 0.01 mol of $$\ce{OH-}$$ is left) when there is also a backward reaction producing $$\ce{OH-}$$ and $$\ce{HCN}$$ back.

• You can directly calculate, what is HCN/CN- ratio in 0.17M NaOH, And What would be pH of NaCN alone Consider more thorough elaboration of your effort to avoid closing future questions. – Poutnik May 6 at 6:57

Your doubt is legitimate because you are assuming that $$K_{(a)}$$ is a parameter that you can plug in every reaction. Actually $$K_{(a)}$$ is calculated using this equation in a solution of the acid with water:

$$\ce{K_{(a)} = \frac{[H+][CN-]}{[HCN]}}$$ That is derived from this ideal equilibrium: $$\ce{HCN ⇌ CN- + H+}$$

Even though it reflects the strength of that specific acid in a pure water solution it's practical usage is limited when the equilibrium is not perturbated by others factors. Anything that will interfere with this equilibrium can cause problems to your assumptions.

In your specific case, the conditions are different there is an excess of $$\ce{OH-}$$ in the solution due to the strong base, so every $$\ce{H+}$$ produced will react with the $$\ce{OH-}$$. Due to the Le Chatelier principle the equilibrium will respond shifting to the right: dissociating more and more $$\ce{HCN}$$ from producing more $$\ce{H+}$$.

In simple words, if one reaction is favourable the solution doesn't care about your $$K_{(a)}$$ it will go following the most thermodynamic favourable path to the equilibrium.

## Possible practical learning aim of this question?

So if you want to have high products amount (i.e. $$\ce{CN-}$$) and you have cheap reactant (i.e. $$\ce{KOH}$$) you can use an excess of reactant to shift the equilibrium where you want.

## Bonus question: what can you do if you want to have HCN?

• Really good answer, your explanation of the equilibrium shifting to the right in a continuous cycle until no more HCN is left clarified my doubt in an instant. Thanks for your help. – Shashwat Tomar May 6 at 20:45