Chemistry

Problem

A laboratory has a stock solution of a strong monoprotic acid with a measured pH of

1.50

25^\circ\text{C}

. A technician dilutes

10.00\,\text{mL}

of this stock acid to a final volume of

500.0\,\text{mL}

using deionized water. (a) Calculate the hydronium-ion concentration in the original stock solution. (b) Determine the hydronium-ion concentration in the diluted solution after the

10.00\,\text{mL}

aliquot is brought to

500.0\,\text{mL}

total volume. Assume volumes are additive. (c) Compute the pH of the diluted solution, expressing the result with the correct number of significant figures. (d) For a

250.0\,\text{mL}

portion of the diluted solution, calculate the mass (in grams) of hydrogen ions present. Express your answer in scientific notation.

Intuitive explanation

The key here is understanding how dilution affects the concentration of $\text{H}^+$ ions in a strong acid. Start by converting the initial pH to find the original $[\text{H}^+]$ and molarity, then use the dilution relationship $M_1V_1 = M_2V_2$ to find the new concentration. Finally, convert that diluted concentration back to pH—this shows why diluting a strong acid by a factor of 50 doesn't simply divide the pH by 50, but instead shifts it by a predictable amount based on logarithmic scale.

Intuitive explanation

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