Question

Acid - base titrationsWhat volume of 0.675 M HCl is required to neutralize 20.0 mL of 0.328 M Ba(OH)2 solution? Balance thee equation for the reaction:HCl + Ba(OH)2 arrow BaCl2 + H2O

Answer 1

The volume of acid used to neutralize the base is 19.43 mL

Step-by-step explanation

Given that;

The volume of Ba(OH)2 used is 20mL

The concentration of Ba(OH)2 is 0.328 M

The concentration of HCl used is 0.675M

To find the volume of HCl used, follow the steps below

Step 1; Write a balanced chemical equation for the reaction

`\text{ 2HCl }+\text{ Ba\lparen OH\rparen}_2\text{ }\rightarrow\text{ BaCl}_2\text{ }+\text{ 2H}_2O`

In the above equation, 2 moles HCl react with 1 mole Ba(OH)2 to produced 1 mole BaCl2 and 2 moles H2O

Step 2; Write the acid-base titration formula

`\text{ }(C_AV_A)/(C_BV_B)\text{ }=\text{ }(n_A)/(n_B)`

Where

CA is the concentration of the acid (HCl)

VA is the volume of the acid (HCl)

CB is the concentration of the base (Ba(OH)2

VB is the volume of the base is (Ba(OH)2

nA is the number of moles of acid

nB is the number of moles of base

Step 3; Substitute the given data into the formula in step 2

`\begin{gathered} \text{ }\frac{0.675* VA}{0.328*\text{ 20}}=(2)/(1) \\ \text{ Cross multiply} \\ \text{ 0.675 }*\text{ V}_A\text{ }*1=\text{ 2 }*\text{ 0.328}*\text{ 20} \\ \text{ 0.675V}_A\text{ }=13.12 \\ \text{ Divide both sides by 0.675} \\ \text{ V}_A\text{ }=\text{ }(13.12)/(0.675) \\ \text{ V}_A=19.43\text{ mL} \end{gathered}`

Hence, the volume of acid used to neutralize the base is 19.43 mL