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Calculate ph of a 0.150m solution of malonic acid. Ionization constant for the acid are K(1)= 1.40 x 10 ^ -3 ; K(2) = 2.2 x 10^-6 give the figure of the answer using first ionization constant now what is the ph for the Solution containing ha ion using the given K(2)
Accepted Answer
Step 1: Calculate pH using the first ionization constant (K1)
Malonic acid (H2C3H2O4) is a diprotic acid. We'll first consider the first ionization:
H2C3H2O4(aq) ⇌ HC3H2O4-(aq) + H+(aq)
Let 'x' be the concentration of H+ ions at equilibrium. We can set up an ICE table:
| | H2C3H2O4 | HC3H2O4- | H+ |
|---|---|---|---|
| I | 0.150 | 0 | 0 |
| C | -x | +x | +x |
| E | 0.150-x | x | x |
We can use the K1 expression:
K1 = [HC3H2O4-][H+]/[H2C3H2O4]
1.40 x 10^-3 = x^2 / (0.150 - x)
Since K1 is relatively small, we can approximate (0.150 - x) ≈ 0.150
1.40 x 10^-3 = x^2 / 0.150
x^2 = 2.1 x 10^-4
x = [H+] = 0.0145 M
Therefore, pH = -log[H+] = -log(0.0145) = 1.84
Step 2: Calculate pH for the solution containing HA- ion using K2
Now we consider the second ionization:
HC3H2O4-(aq) ⇌ C3H2O4^2-(aq) + H+(aq)
We'll use the initial concentration of HC3H2O4- as approximately equal to the H+ concentration from the first ionization (0.0145 M). Let 'y' be the concentration of H+ ions from the second ionization.
| | HC3H2O4- | C3H2O4^2- | H+ |
|---|---|---|---|
| I | 0.0145 | 0 | 0.0145 |
| C | -y | +y | +y |
| E | 0.0145-y | y | 0.0145+y |
K2 = [C3H2O4^2-][H+]/[HC3H2O4-]
2.2 x 10^-6 = y(0.0145+y) / (0.0145 - y)
Since K2 is very small, we can approximate (0.0145 + y) ≈ 0.0145 and (0.0145 - y) ≈ 0.0145
2.2 x 10^-6 = y(0.0145) / 0.0145
y = [H+] = 2.2 x 10^-6 M
Therefore, pH = -log[H+] = -log(2.2 x 10^-6) = 5.66
Malonic acid (H2C3H2O4) is a diprotic acid. We'll first consider the first ionization:
H2C3H2O4(aq) ⇌ HC3H2O4-(aq) + H+(aq)
Let 'x' be the concentration of H+ ions at equilibrium. We can set up an ICE table:
| | H2C3H2O4 | HC3H2O4- | H+ |
|---|---|---|---|
| I | 0.150 | 0 | 0 |
| C | -x | +x | +x |
| E | 0.150-x | x | x |
We can use the K1 expression:
K1 = [HC3H2O4-][H+]/[H2C3H2O4]
1.40 x 10^-3 = x^2 / (0.150 - x)
Since K1 is relatively small, we can approximate (0.150 - x) ≈ 0.150
1.40 x 10^-3 = x^2 / 0.150
x^2 = 2.1 x 10^-4
x = [H+] = 0.0145 M
Therefore, pH = -log[H+] = -log(0.0145) = 1.84
Step 2: Calculate pH for the solution containing HA- ion using K2
Now we consider the second ionization:
HC3H2O4-(aq) ⇌ C3H2O4^2-(aq) + H+(aq)
We'll use the initial concentration of HC3H2O4- as approximately equal to the H+ concentration from the first ionization (0.0145 M). Let 'y' be the concentration of H+ ions from the second ionization.
| | HC3H2O4- | C3H2O4^2- | H+ |
|---|---|---|---|
| I | 0.0145 | 0 | 0.0145 |
| C | -y | +y | +y |
| E | 0.0145-y | y | 0.0145+y |
K2 = [C3H2O4^2-][H+]/[HC3H2O4-]
2.2 x 10^-6 = y(0.0145+y) / (0.0145 - y)
Since K2 is very small, we can approximate (0.0145 + y) ≈ 0.0145 and (0.0145 - y) ≈ 0.0145
2.2 x 10^-6 = y(0.0145) / 0.0145
y = [H+] = 2.2 x 10^-6 M
Therefore, pH = -log[H+] = -log(2.2 x 10^-6) = 5.66
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