Question

Calculate the pH of each of the following strong acid solutions.(a) 0.00327 M HIpH =(b) 0.649 g of HCl in 46.0 L of solutionpH =(c) 67.0 mL of 7.30 M HI diluted to 1.60 LpH =(d) a mixture formed by adding 19.0 mL of 0.00246 M HI to 44.0 mL of 0.00871 M HClpH =

Answer 1

Step-by-step explanation:

(a) The pH of 0.00327 M HI can be calculated as follows:

HI → H+ + I-

Since HI is a strong acid, it dissociates completely in water, giving H+ ions. Thus, the [H+] concentration is equal to the initial acid concentration:

[H+] = 0.00327 M

The pH can be calculated as:

pH = -log[H+]

= -log(0.00327)

= 2.484

Therefore, the pH of 0.00327 M HI is 2.484.

(b) The number of moles of HCl in 0.649 g can be calculated as follows:

molar mass of HCl = 1 g/mol + 35.5 g/mol = 36.5 g/mol

moles of HCl = mass/molar mass = 0.649 g/36.5 g/mol = 0.0178 mol

The molarity of the HCl solution is:

Molarity = moles/volume = 0.0178 mol/46.0 L = 0.000387 M

Since HCl is a strong acid, it dissociates completely in water, giving H+ ions. Thus, the [H+] concentration is equal to the initial acid concentration:

[H+] = 0.000387 M

The pH can be calculated as:

pH = -log[H+]

= -log(0.000387)

= 3.412

Therefore, the pH of 0.649 g of HCl in 46.0 L of solution is 3.412.

(c) The initial number of moles of HI is:

moles of HI = concentration × volume = 7.30 M × 0.0670 L = 0.4891 mol

After dilution, the final concentration of HI is:

final concentration = initial concentration × (initial volume/final volume)

= 7.30 M × (0.0670 L/1.60 L)

= 0.306 M

Since HI is a strong acid, it dissociates completely in water, giving H+ ions. Thus, the [H+] concentration is equal to the initial acid concentration:

[H+] = 0.306 M

The pH can be calculated as:

pH = -log[H+]

= -log(0.306)

= 0.513

Therefore, the pH of the diluted solution is 0.513.

(d) The number of moles of HI and HCl in the mixture can be calculated as follows:

moles of HI = concentration × volume = 0.00246 M × 0.0190 L = 4.674 × 10^-5 mol

moles of HCl = concentration × volume = 0.00871 M × 0.0440 L = 0.000383 mol

The total volume of the mixture is:

total volume = 0.0190 L + 0.0440 L = 0.0630 L

The total number of moles of acid in the mixture is:

total moles of acid = moles of HI + moles of HCl

= 4.674 × 10^-5 mol + 0.000383 mol

= 0.00043 mol

The molarity of the acid mixture is:

Molarity = total moles of acid/total volume = 0.00043 mol/0.0630 L = 0.00683 M

Since the mixture contains both HI and HCl, the pH cannot be calculated directly. However, since both are strong acids, they will dissociate completely