Question

A patient is suspected of having low stomach acid, a condition known as hypochloridia. To determine whether the patient has this condition, her doctors take a 19.00 mL sample of her gastric juices and titrate the sample with 0.000300 M KOH. The gastric juice sample required 12.1 mL of the KOH titrant to neutralize it. Calculate the pH of the gastric juice sample. Assume the sample contained no ingested food or drink which might otherwise interfere with the titration.

Answer 1

Concepts and reason

The problem is based on the concept of the pH of an acid can be calculated by titrating it against a base of known concentration as the gastric juice in human stomach is acidic in nature because it contains ${\rm{HCl}}$ .

Fundamentals

The principle of titration is that number of moles of acid $\left( {{n_{\rm{A}}}} \right)$ react with equal number of moles $\left( {{n_{\rm{B}}}} \right)$ of base.

${n_{\rm{A}}} = {n_{\rm{B}}}$ …… (1)

The pH of an acidic solution is

${\rm{pH}} = - \log \left[ {{{\rm{H}}^ + }} \right]$ …… (2)

Here, $\left[ {{{\rm{H}}^ + }} \right]$ is concentration of hydrogen ion.

Molarity (M) is defined as the number of moles of solute distributed in per litre of a solution.

$M = {\rm{ }}\frac{{n\left( {{\rm{moles}}} \right)}}{{V\left( {\rm{L}} \right)}}$ …… (3)

Here $n$ is the number of moles and $V$ is the volume of the solution in liters.

Number of moles $\left( n \right)$ in a solution is the product of molarity (M) and volume of the solution (V). Rearranging equation (3) this equation is obtained.

$n = M \times V$ …… (4)

Molarity of ${\rm{KOH}}$ solution $\left( {{M_{{\rm{NaOH}}}}} \right)$ is ${\rm{0}}{\rm{.000300 M}}$ and the volume of ${\rm{KOH}}$ used $\left( {{V_{{\rm{KOH}}}}} \right)$ is 12.1 mL

Number of moles of KOH $({n_{KOH}})$ can be determined from equation (2).

Substitute, $0.000300{\rm{ M}}$ for $M$ and $12.1{\rm{ mL}}$ for $V$ in equation (2).

$\begin{array}{c}\\n = 0.000300{\rm{ M}} \times 12.1{\rm{ mL}}\\\\ = \left( {0.000300{\rm{ }}\frac{{{\rm{mol}}}}{{\rm{L}}} \times 12.1{\rm{ mL}} \times \frac{{1{\rm{ L}}}}{{1000{\rm{ mL}}}}} \right)\\\\ = 3.63 \times {10^{ - 6}}{\rm{ mol}}\\\end{array}$

Number of moles of ${\rm{HCl}}$ $\left( {{n_{{\rm{HCl}}}}} \right)$ will be equal to the number of moles of ${\rm{KOH}}$ $\left( {{n_{{\rm{KOH}}}}} \right)$

${n_{{\rm{HCl}}}} = {n_{{\rm{KOH}}}}$

Substitute the value of ${n_{{\rm{KOH}}}}$ in the above expression.

${n_{{\rm{HCl}}}} = 3.63 \times {10^{ - 6}}{\rm{ mol}}$

Molarity of ${\rm{HCl}}$ $\left( {{n_{{\rm{HCl}}}}} \right)$ can be calculated from equation (3).

${M_{{\rm{HCl}}}} = \frac{{{n_{{\rm{HCl}}}}}}{{{V_{{\rm{HCl}}}}}}$

Substitute $3.63 \times {10^{ - 6}}{\rm{ mol}}$ for ${n_{{\rm{HCl}}}}$ and 19.00 mL for ${V_{{\rm{HCl}}}}$ to the above expression.

$\begin{array}{c}\\{M_{{\rm{HCl}}}} = \frac{{3.63 \times {{10}^{ - 6}}{\rm{ mol}}}}{{19.00{\rm{ mL}}}}\\\\ = \frac{{3.63 \times {{10}^{ - 6}}{\rm{ mol}}}}{{19.00{\rm{ mL}}}} \times \frac{{1000{\rm{ mL}}}}{{1{\rm{ L}}}}\\\\ = 1.91 \times {10^{ - 4}}{\rm{ mol }}{{\rm{L}}^{ - 1}}\\\end{array}$

Concentration of ${\rm{HCl}}$ $\left( {{M_{{\rm{HCl}}}}} \right)$ is equal to concentration of ${{\rm{H}}^{\rm{ + }}}$ ions.

$\begin{array}{c}\\\left[ {{{\rm{H}}^ + }} \right] = \left[ {{\rm{HCl}}} \right]\\\\\left[ {{{\rm{H}}^ + }} \right] = {M_{{\rm{HCl}}}}\\\end{array}$

Substitute $1.91 \times {10^{ - 4}}{\rm{ mol }}{{\rm{L}}^{ - 1}}$ for ${M_{{\rm{HCl}}}}$ .

$\left[ {{{\rm{H}}^ + }} \right] = 1.91 \times {10^{ - 4}}{\rm{ mol }}{{\rm{L}}^{ - 1}}$

The pH of the solution can be calculated from the equation (2).

Substitute, $1.91 \times {10^{ - 4}}{\rm{ mol }}{{\rm{L}}^{ - 1}}$ for $\left[ {{{\rm{H}}^{\rm{ + }}}} \right]$ in equation (2).

$\begin{array}{c}\\{\rm{pH}} = - \log \left[ {1.91 \times {{10}^{ - 4}}{\rm{ mol }}{{\rm{L}}^{ - 1}}} \right]\\\\ = 3.72\\\end{array}$

Ans:

The pH of the gastric juice is $3.72$ .