I have a few questions relating to ph and whatnot.. 1) Calculate the pH of 0.001
ID: 816456 • Letter: I
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, Ka=1.52 x 10-5. Answer in 4 significant figures.
4) Calculate the pH of 0.00000M C5H5N solution. Kb=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, Kb = 1.8 105.
Explanation / Answer
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)
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