Question

I have a few questions relating to ph and whatnot..

1) Calculate the pH of 0.0010 M Ca(OH)₂ solution. (to 2 decimal places)

2) Determine the pH of a KOH solution made by mixing 0.251 g KOH with enough water to make 1.00 � 10² mL of solution.

3) Calculate the pH of 0.0171 butanoic acid, K_a=1.52 x 10^-5. Answer in 4 significant figures.

4) Calculate the pH of 0.00000M C₅H₅N solution. K_b=1.5 x 10^-9. Answer in 3 decimal places.

5) Calculate the pKa of a 0.080M HOCl solution with a pH of 3.22.

6) A solution containing NH₃(aq) and NH₄Cl(aq) has a pH of 9.5. What is the [NH₃]/[NH₄⁺] ratio in this solution? For ammonia, K_b = 1.8 � 10^�5.

Answer 1

1. Since Ca(OH)2 is a strong electrolyte, it is dissociated completely when dissolved in water.

Ca(OH)2 ------à Ca²⁺(aq) + 2OH^-

                             0.0010M   2*0.0010M

                             0.0010M    0.0020M

[OH^-] = 0.0020 M

pOH = - log[OH^-] = - log(0.0020) = 2.70

=> pH = 14 – pOH = 14 -2.70 = 11.30 (answer)

2. Given the mass of KOH = 0.251g

Molecular mass of KOH = 56 gmol^-1

Hence moles of KOH = mass / molecular mass = 0.251g / 56 gmol^-1 = 0.00448 mol

Volume of the solution, V = 1.00*10² mL = (1.00*10² mL)*(1L/10³ mL) = 0.10L

Concentration of KOH solution = moles of KOH / V = 0.00448 mol /0.10L

= 0.0448 M

Since KOH is a strong electrolyte, it will be completely dissociated into K⁺ and OH^-.

Hence [OH^-] = [KOH] = 0.0448 M

pOH = - log[OH^-] = - log(0.0448) = 1.35

Hence pH = 14 – pOH = 14 – 1.35 = 12.65 (answer)

3. Butanoic acid(C3H7COOH) is a weak acid and not dissociated cmpletely.

Given concentration of C3H7COOH, C = 0.0171M

the dissociation of C3H7COOH can be written as

          C3H7COOH -------> C3H7COO^-(aq) + H⁺ (aq) , Ka = 1.52*10^-5

init: 0.0171 M 0M 0M

at eqm :(0.0171 - x) M    x M        x M

Equilibrium constant, Ka for the above reaction can be calculated as

Ka = [CH3COO^-(aq)] * [H⁺ (aq)] / [ CH3COOH]

=> 1.8*10^-5  = x² /(0.0171 - x)

Since x << 0.0171 , 0.0171 – x is nearly equals to 0.0171.

Hence  1.52*10^-5  = x² /0.0171

=> x = underroot(0.0171*1.52*10^-5) = 5.10*10^-4

pH = - log[H+] =- log(5.10*10^-4) = 3.292 (answer)