Chapter 9.6, Problem 15PPB

What volume (in mL) of a 0.2550 M NaOH solution can be neutralized by 10.75 g of KHP?

Interpretation Introduction

Interpretation:

The volume of $NaOH$ needed to neutralize the given mass of $KHP$ should be calculated

Concept introduction:

Neutralization reaction:

• The reaction of acid and base to gives salt is known as neutralization reaction.
• The equal volume and concentrations of acid is equally react with equal volume and concentrated base.
• In end point the ionic concentration is zero.

Mole:

Mole of the solute in solution is given by multiplication of concentration (molarity) with volume.

$Mole=concentration\left(M\right)\times Volume(L)$

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mass:

The Mass of compound is given by the multiplication of mole of compound with molar mass of compound.

$\text{Massof}\text{the}\text{compound}=\text{Mole}\times \text{Molar}\text{mass}\text{of}\text{the}\text{compound}$

Answer to Problem 15PPB

• The volume of $NaOH$ needed to neutralize the given mass of $KHP$ is $\text{2245}\text{mL}$.

Explanation of Solution

The volume of $NaOH$ needed to neutralize the given mass of $KHP$

The balanced equation is,

${\text{KHC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{}_{\text{(aq)}}{\text{+NaOH}}_{\text{(aq)}}^{}\to {\text{HC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}{\text{Na}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Molar mass of ${\text{KHC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$ is $187.77g$

Mole of ${\text{KHC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$ solution is,

$\begin{array}{l}\text{=}\frac{\text{10}\text{.75 g}}{\text{187}\text{.77}\text{g}}\\ \text{=}\text{0}\text{.05725}\text{moles}\end{array}$

• The mass and molar mass of ${\text{KHC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$ are plugged in above equation to gives a mole of ${\text{KHC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$ in taken gram of sample.

  $\begin{array}{l}\text{Volume of NaOH}\text{=}\frac{\text{0}\text{.5725}}{\text{0}\text{.2550}\text{M}}\\ \text{=}\text{2}\text{.245}\text{L}\\ \text{=}\text{2245}\text{mL}\end{array}$

• The $NaOH$ is requires a equal mole of ${\text{KHC}}_{\text{8}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$ to neutralized.  Therefore the mole of $NaOH$ equal to required mole of $KHP$ to complete neutralization.
• Hence the calculated mole and molarity of $NaOH$ are plugged in the equation to give volume of $NaOH$ needed to neutralize the given $KHP$ solution.
• The volume of $NaOH$ needed to neutralize the given mass of $KHP$ is $\text{2245}\text{mL}$

Conclusion

The volume of $NaOH$ needed to neutralize the given mass of $KHP$ was calculated