x = squareroot (C_0 * K_a) was a result from the weak acid reaction with water e
ID: 498802 • Letter: X
Question
x = squareroot (C_0 * K_a) was a result from the weak acid reaction with water equilibrium. HA(aq) + H^0_2O(l)^a A^-(aq) + H_3O^+ (aq). What is x equal to? A the change in concentration of HA B the change in concentration of A^- C the change in concentration of H_3O+ D the pH 2. What is 1.5 times 10^-3 + 1.0times 10^-7? A 1.0 times 10^-7 B 0 C 1.5 times10^-3 1.5001 times 10^-3 The change in initial concentrations to reach equilibrium is always A positive for products and negative for reactants B positive for reactants and negative for products c positive for products and positive for reactants D negative for products and negative for reactants x = squareroot (C_0 * K_a) was a result from the weak acid reaction with water equilibrium. HA (aq) + H^0_2O(l)^a A^-(aq) + H_3O^+(aq). What is the pH of a solution of acetic acid with a concentration of 0.025 M and a pK_a = 4.8. A 2.2 B 3.2 C 5.2 D 10.8 HA(ag) + H_2O(I) A^-(aq) + H_3O^+ (aq). Addition of which quaExplanation / Answer
1) HA(aq) + H2O (l) -------------> H3O+(aq) + A- (aq)
Co 0 0
Co-x x x
Then,
Ka = [H3O+] [A-]/ [HA]
= x.x/ (Co-x)
Since x is very small, Co-x = Co
Hence,
Ka = x.x/ Co
x2 = Co.Ka
x = (Co*Ka)1/2
x is equal to change in concentration of H3O+
Therefore,
ans = C
2)
1.5 x 10-3 + 1.0 x 10-7
= 10-3 [ 1.5 + 10-4]
= 10-3 [ 1.5 + 0.0001]
= 1.5001 x 10-3
Therefore,
ans = D
3)
The change in initial concentrations to reach equilibrium is always - negative for reactants and positive for products.
Therefore,
ans = A
4)
[H3O+] = [H+] = (Co*Ka)1/2
pH = - log [H+] = -log (Co*Ka)1/2
= [-logCo - logKa]/2
= [pKa - logCo]/2
= [4.8 - log 0.025]/2
= 3.2
pH = 3.2
Therefore,
ans = B
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