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

To calculate the pH of a solution made from hydrochloric acid or sodium hydroxide, you need to follow certain steps. In both cases, you need to calculate the concentration of the respective ions and then use the pH formula to find the pH.

Step-by-step explanation:

The pH of a solution can be calculated using the formula pH = -log[H+]. In the case of hydrochloric acid (HCl) solution made by dissolving 0.35 g of HCl in 2.5 L of water, we need to first calculate the molarity of HCl by dividing the number of moles of HCl by the volume of the solution. Then, we can use this molarity value to calculate the hydrogen ion concentration [H+], and finally, find the pH using the pH formula.

For the sodium hydroxide (NaOH) solution made by dissolving 0.80 g of NaOH in 2.0 L of water, we can follow the same steps as before to calculate the pOH of the solution. Once we have the pOH, we can use the relationship pH + pOH = 14 to find the pH.

Answer 2

• A) pH = 2.42
• B) pH = 12.00

Step-by-step 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)*(1mol_(HCl))/(36.46g_(HCl)) *(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)*(1mol_(NaOH))/(40g_(NaOH)) *(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