Question

1a. Calculate the pH of a 8.01 × 10⁻⁵M HBr solution.

Report your answer to TWO places past the decimal.

1b. . Calculate the pH of a 1.39 × 10⁻⁶M HNO₃ solution.

Report your answer to TWO places past the decimal.

1c. Calculate the pH of a 0.00963 M Ca(OH)₂ solution.

Report your answer to TWO places past the decimal.

1d. Calculate the pH of a 0.00155 M Ca(OH)₂ solution.

Report your answer to TWO places past the decimal.

1e. Calculate the pH of a solution that contains 0.0705 g of Ba(OH)₂ in 1.65 L of water.

Report your answer to TWO places past the decimal.

1f. Calculate the pH of a solution that contains 0.0243 g of Mg(OH)₂ in 2.11 L of water.

Report your answer to TWO places past the decimal.

please answer all parts and show your work thank you so much!!

Answer 1

(1a): HBr is strong acid , it will dissociate completely in the solution . [HBr] = [H⁺] = 8.01×10^-5M  

pH = - log[H⁺] = 5 - log(8.01) = 4.096 = 4.10 (answer)

(1b) : concentration of HNO₃ is very small, we need to consider the concentration from water also.

[H⁺] = (1.39×10^-6 + 10^-7 )M = 1.49×10^-6

pH = - log(1.49×10^-6) = 5.827 = 5.83 (answer)

(1c) : Concentration of Ca (OH)₂ = 0.00963M ,

Concentration of OH^- = 2 × 0.00963M = 0.01926M

pOH = - log(0.01926) = 1.7153

pH = 14 - pOH = 14 - 1.7153 = 12.2847 = 12.18 (answer)

(1d) : [OH^-] = 2 × 0.00155M = 0.00310

pOH = - log(0.00310) = 2.5086

pH = 14 - pOH = 14 - 2.5086 = 11.4914 = 11.49 (answer)

(1e) : molarity = (mass of substance /molar mass )/volume of solution(L)

molarity of Ba(OH)₂ = (0.0705g / 171.342g/mol)/1.65L = 2.4937×10^-4M

[OH^-] = 2×2.4937×10^-4M

pOH = 3.302

pH = 14 - pOH = 14 - 3.302 = 10.698 = 10.70 (answer)