Titration-like problem

Question

What quantity of NaOH(s) must be added to 1.00 L of 0.492 M HCL to achieve a pH of 13.00? (assume no volume change)

 

Answer 1

You need to add enough NaOH to first neutralize the HCl, and then add the NaOH to make pH 13. HCl is a strong acid and will dissociate completely. NaOH is a strong base and will dissociate completely. This means:

 

HCl → H⁺ + Cl^-

and

NaOH → Na⁺ + OH^-

 

Moles of HCl = moles of H⁺ 

Moles of NaOH = moles OH^-

 

To neutralize the acid, we need to see the ratio the acid and the base react:

NaOH + HCl → NaCl + H₂O

 

We see the acid and the base react in a ratio of 1:1. To neutralize the acid, we find out how many moles of acid we start with and add equal amounts of base:

 

(0.492 moles HCl/1 L) x 1.00 L = 0.492 moles HCl

0.492 moles HCl x (1 mol H⁺)/(1 mol HCl) = 0.492 moles H⁺

 

We must add 0.492 moles of NaOH to neutralize the acid. Now we are at pH of 7 (neutral). We then find out how much NaOH to add to get the pH to 13. This is the same thing as getting the pOH to 1, since pH + pOH = 14. We use the equation:

 

-log[OH^-] = pOH

We want pOH = 1

-log[OH-] = 1

Now solve for [OH^-]

[OH^-] = 10^-1 = 0.1 M OH^- = 0.1 M NaOH

 

Since there is no volume change, we are adding an additional 0.1 moles to the already added base. Therefore, the quantity of NaOH_(s) is:

 

0.492 moles NaOH + 0.1 moles NaOH = 0.592 moles NaOH.