Question

A sample of 25.0 mL of 0.120 M Ca(OH)2(aq) is titrated with 0.150 M HCl(aq). What volume of HCl(aq) is needed to completely neutralize the Ca(OH)2(aq)

Answer 1

Answer:

$V_{HCl}=48.0 mL$

Explanation:

Hello!

In this case, since the reaction between calcium hydroxide and hydrochloric acid shows a 2:1 mole ratio between them:

$2HCl+Ca(OH)_2\rightarrow CaCl_2+2H_2O$

We need to use the following mole ratio:

$2n_{Ca(OH)_2}=n_{HCl}$

That can be written in volumes and concentrations:

$2M_{Ca(OH)_2}V_{Ca(OH)_2}=M_{HCl}V_{HCl}$

Thus, we solve for the volume of HCl as it is the unknown:

$V_{HCl}=\frac{2M_{Ca(OH)_2}V_{Ca(OH)_2}}{M_{HCl}}$

Therefore, we plug in to obtain:

$V_{HCl}=\frac{2*0.120M*25.0mL}{0.150M} \\\\V_{HCl}=48.0 mL$

Best regards!