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 $\ce{HCN + OH- ⇌ CN- + H2O}$


4. So 0.062-0.052 mol=0.01 mol of $\ce{OH-}$ is left over
5. 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.


8. 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. 


Please help.




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](https://en.wikipedia.org/wiki/Le_Chatelier%27s_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?
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