Question

How many grams of K O H are needed to neutralize 10.8 mL of 0.13 M H C l in stomach acid?

Answer 1

Write the reaction as foll ows:

\(\mathrm{KOH}+\mathrm{HCl} \longrightarrow \mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}\)

1 mole of \(\mathrm{KOH}\) reacts with \(1 \mathrm{~mol}\) of \(\mathrm{HCl}\) and forms \(1 \mathrm{~mole}\) of \(\mathrm{KCL}\) and \(1 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\).

Volume of \(\mathrm{HCl}=10.8 \mathrm{~mL}\)

molarity of \(\mathrm{HCl}=0.13 \mathrm{M}\)

Calculate the moles of \(\mathrm{HCl}\) as foll ows moles of \(\mathrm{HCl}=\) molarity of \(\mathrm{HCl} \times\) volume of \(\mathrm{HCl}\)

$$ \begin{aligned} &=0.13 \mathrm{M} \times \frac{\mathrm{mol} / \mathrm{L}}{\mathrm{M}} \times 10.8 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} \\ &=0.001404 \mathrm{~mol} \end{aligned} $$

From the reaction, moles of \(\mathrm{KOH}=\) moles of \(\mathrm{HCl}\) moles of \(\mathrm{KOH}=0.001404 \mathrm{~mol}\)

\(-3\)

molar mass of \(\mathrm{KOH}=56.1 \mathrm{~g} / \mathrm{mol}\)

Calcuate the mass of \(\mathrm{KOH}\) as follows:

mass of \(\mathrm{KOH}=\) molar mass of \(\mathrm{KOH} \times \mathrm{moles}\) of \(\mathrm{KOH}\) mass of \(\mathrm{KOH}=56.1 \mathrm{~g} / \mathrm{mol} \times 0.001404 \mathrm{~mol}\)

mass of \(\mathrm{KOH}=0.0788 \mathrm{~g}\)

mass of \(\mathrm{KOH}=78.8 \times 10^{-3} \mathrm{~g}\)

Therefore, the required solution is \(78.8 \times 10^{-3} \mathrm{~g}\)