Question

In an acid-base neutralization reaction 38.74 ml of 0.500 m potassium hydroxide (ki) reacts with 50.00 ml of sulfuric acid solution. what is the concentration of the h2so4 solution?

Answer 1

Answer: 0.19M

Explanation: $2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O$

Using Molarity equation:

$n_1M_1V_1=n_2M_2V_2$

$M_1$ = molarity of acid

$V_1$ = volume of acid

$M_2$ = molarity of base

$V_2$ = volume of base

$2\times M_1\times 50ml=1\times 0.500M\times 38.74ml$

$M_1=0.19M$

Thus the concentration of the $H_2SO_4$ solution is 0.19M.

Answer 2

Concentration of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ solution is $\boxed{{\mathbf{0}}{\mathbf{.1937 mol}}}$ .

Further Explanation:

Neutralization reaction:

It is the reaction that occurs between an acid and a base in order to form salt and water. It is named so as it neutralizes the excess amount of hydrogen or hydroxide ions present in the solution. An example of neutralization reaction is,

${\text{HCl}}+{\text{NaOH}}\to {\text{NaCl}}+{{\text{H}}_2}{\text{O}}$

The balanced chemical equation for the neutralization reaction of potassium hydroxide and sulphuric acid solution is as follows:

${\text{2KOH}}\left({aq}\right)+{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\left( {aq}\right)\to {{\text{K}}_2}{\text{S}}{{\text{O}}_4}\left({aq}\right)+2{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right)$

From the balanced chemical reaction, the reaction stoichiometry between  ${\text{KOH}}$ and ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is as follows:

$\boxed{2{\text{ mol KOH}}:1{\text{ mol }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}$

The balanced chemical equation shows that 2 moles of ${\text{KOH}}$ reacts with 1 mole of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ to neutralize the reaction completely.

The volume of given solution of ${\text{KOH}}$ is 38.74 ml and the concentration of ${\text{KOH}}$ is 0.500 M

Calculate the number of moles of  ${\mathbf{KOH}}$ as follows:

$\begin{aligned}{\text{Number of moles of KOH}}\left({{\text{mol}}}\right)&= {\text{Concentration }}\left({{\text{mol/L}}}\right)\times{\text{Volume }}\left( {\text{L}}\right)\\&=0.500{\text{ mol/L}}\times\left({38.74{\text{ ml}}\times\frac{{1{\text{ L}}}}{{1000\,{\text{ml}}}}}\right)\\&=0.01937{\text{ mol}}\\\end{gathered}$

Since 2 moles of ${\text{KOH}}$ reacts with1 mole of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ to neutralize the reaction completely, therefore, the number of moles of ${{\mathbf{H}}_{\mathbf{2}}}{\mathbf{S}}{{\mathbf{O}}_{\mathbf{4}}}$  required to neutralize 0.01937 moles of ${\text{KOH}}$ is calculated as follows:

$\begin{aligned}{\text{Amount of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\left( {{\text{mol}}}\right)&=\left({{\text{0}}{\text{.01937 mol KOH}}} \right)\left({\frac{{1{\text{ mol }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}}{{2{\text{ mol KOH}}}}} \right)\\&= 0.009685{\text{ mol }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\\\end{gathered}$

Given volume of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is 50 ml.

The concentration of ${{\mathbf{H}}_{\mathbf{2}}}{\mathbf{S}}{{\mathbf{O}}_{\mathbf{4}}}$ can be calculated as follows:

$\begin{aligned}{\text{Concentration }}\left( {{\text{mol/L}}}\right)&= \frac{{{\text{Number of moles of }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({{\text{Mol}}}\right)}}{{{\text{Volume}}\left( {\text{L}}\right)}}\\&=\frac{{0.009685{\text{ mol}}}}{{{\text{50 ml}}\times\left({\frac{{1\;{\text{L}}}}{{1000\;{\text{ml}}}}}\right)}}\\&=0.1937{\text{ mol}}\\\end{gathered}$

Learn more:

1. Determine the molarity of the h3po4 solution: brainly.com/question/9850829

2. The concentration of molecules during diffusion: brainly.com/question/1386629

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical reactions and equations

Keywords: Concentrations, balance chemical equation, neutralization reaction,potassium hydroxide, KOH, sulphuric acid, H2SO4, stoichiometry, number of moles.