Question

A 100.0 mL sample of 0.18 M HClO4 (Strong acid) is titrated with 0.27 M LiOH. Determine the pH of the solution after the addition of 75.0 mL of LiOH

Answer 1

Answer:

pH = 12.11

Explanation:

Hello!

In this case, since the chemical reaction between perchloric acid and lithium hydroxide is:

$HClO_4+LiOH\rightarrow LiClO_4+H_2O$

Whereas the mole ratio between the acid and base is 1:1, it means they react in the same proportion. In such a way, given the volume and concentration of acid, we compute the moles:

$n_{acid}=0.1000L*0.18mol/L=0.018mol$

Now, the moles of base that reacted:

$n_{base}=0.075L*0.27mol/L=0.02025mol$

Thus, since there is an excess of lithium hydroxide of:

$n_{LiOH}^{remaining}=0.02025-0.018=0.00225mol$

We compute the new concentration considering the total final volume of 175.0 mL:

$[LiOH]=\frac{0.00225mol}{0.1750L}=0.013M$

Now, since there is a strong base remaining in solution, we compute its pOH first:

$pOH=-log([OH^-])=-log(0.013M)=1.89$

Because LiOH is fully ionized to Li⁺ and OH⁻ ions. Therefore the pH is:

$pH+pOH=14\\\\pH=14-pOH=14-1.89\\\\pH=12.11$

Best regards!