Question

How many grams of HI should be added to \(265 \mathrm{~mL}\) of \(0.215 \mathrm{M} \mathrm{HCl}\) so that the resulting solution has a \(\mathrm{pH}\) of \(0.38 ?\) Assume that the addition of HI does not change the volume of the resulting solution.

Short answer

Answer: Approximately 6.84 grams of HI should be added.

Step-by-step solution

Calculate the [H+] concentration of the resulting solution

We are given the desired pH value, which is 0.38. We can use the formula of pH to find the [H+] concentration: pH = -log([H+]) Now, we need to find the [H+] concentration for pH 0.38: [H+] = 10^(-pH) = 10^(-0.38) = 0.417 M (approx)

Calculate the [H+] concentration of HCl solution

The given solution of HCl is 0.215 M. Since HCl is a strong acid, it completely dissociates in water, so the concentration of H+ ions is equal to the molarity of HCl: [H+]_HCl = 0.215 M

Calculate the [H+] concentration of HI in the mixture

We know the total [H+] concentration from the resulting solution, and we have the [H+] concentration of HCl. So, we can find the [H+] concentration of HI needed in the mixture as follows: [H+]_HI = [H+]_total - [H+]_HCl = 0.417 - 0.215 = 0.202 M

Find the number of moles of HI required

The total volume of the resulting solution is 265 mL (given). To find the moles of HI needed, we will use the formula: moles = Molarity x Volume Let's convert the volume from mL to Liters: Volume = 265 mL / 1000 = 0.265 L Now, we will use the formula to find the moles of HI required: moles_HI = [H+]_HI x Volume = 0.202 M x 0.265 L = 0.05353 moles (approx)

Convert moles of HI to grams

To find the mass (in grams) of the HI needed, we will use the formula: Mass = moles x molar_mass The molar mass of HI = 127.91 g/mol (1.007 + 126.9 for H and I respectively) Now, we will use the formula to find the mass of HI required: Mass_HI = moles_HI × molar_mass = 0.05353 × 127.91 g/mol ≈ 6.84 g (approx) So, approximately 6.84 grams of HI should be added to the 265 mL of 0.215 M HCl solution to achieve a pH of 0.38.