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

pH = 12.11

Step-by-step 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]=(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!