Question

PH is a logarithmic scale used to indicate the hydrogen ion concentration, [H+], of a solution: pH=−log[H+] Due to the autoionization of water, in any aqueous solution, the hydrogen ion concentration and the hydroxide ion concentration, [OH−], are related to each other by the Kw of water: Kw=[H+][OH−]=1.00×10^−14 where 1.00×10^−14 is the value at approximately 297 K. Based on this relation, the pH and pOH are also related to each other as 14.00=pH+pOH. The temperature for each solution is carried out at approximately 297 K where Kw=1.00×10^−14. Part A) 0.35 g of hydrogen chloride (HCl) is dissolved in water to make 2.5 L of solution. What is the pH of the resulting hydrochloric acid solution? Express the pH numerically to two decimal places.
Part B) 0.80 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 2.0 L of solution. What is the pH of this solution?
Express the pH numerically to two decimal places.

Answer 1

Answer:

• A) pH = 2.42
• B) pH = 12.00

Explanation:

The dissolution of HCl is  HCl → H⁺ + Cl⁻

• To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.

The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:

$0.35g_{HCl}*\frac{1mol_{HCl}}{36.46g_{HCl}} *\frac{1molH^{+}}{1mol HCl}$ = 9.60*10⁻³ mol H⁺

So the concentration of H⁺ is

[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M

pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42

• The dissolution of NaOH is  NaOH → Na⁺ + OH⁻
• Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:

$0.80g_{NaOH}*\frac{1mol_{NaOH}}{40g_{NaOH}} *\frac{1molOH^{-}}{1mol NaOH}$= 0.02 mol OH⁻

[OH⁻] = 0.02 mol / 2.0 L = 0.01

pOH = -log [OH⁻] = -log (0.01) = 2.00

With the pOH, we can calculate the pH:

pH + pOH = 14.00

pH + 2.00 = 14.00

pH = 12.00