Question

Can you please help solve and show the calculations so that I can understand?How many milliliters of 0.100M HClO3 are required to neutralize 31.8 mL of 0.140M KOH?

Answer 1

The volume of HClO3 in milliliters required = 44.52 mL

Step-by-step explanation

Given parameters:

Molarity of HClO3, Ca = 0.100 M

Volume of KOH, Vb = 31.8 mL

Molarity of KOH, Cb = 0.140 M

What to find:

The volume, Va of HClO3 in milliliters required.

Step-by-step solution:

Step 1; Write a balanced chemical equation for the reaction.

HClO3 + KOH -----> KClO3 + H2O

Mole ratio of HClO3 to KOH is 1 : 1, that is Na = 1 and Nb = 1

Step 2: Calculate the volume of HClO3 required.

The volume of HClO3 can be calculated using

`\begin{gathered} (C_aV_a)/(n_a)=(C_bV_b)/(n_b) \\ \\ \Rightarrow V_a=(C_bV_bn_a)/(V_an_b) \end{gathered}`

Putting the values of the given parameters into the formula above, we have

`V_a=(0.140M*31.8mL*1)/(0.100M*1)=44.52\text{ }mL`

The volume, Va of HClO3 in milliliters required = 44.52 mL