Calculate the pH of an aqueous solution containing 1.6 x 10-2 M HCl, 2.0 x 10-2 M H2SO4, and 1.8 x 10-2 M HCN.
This is not nearly as easy as it looks. The simple way to do it is to add H^+ from 0.016M HCl, 0.02M H2SO4(first H^+ ONLY) and ignore the H^+ from k2 for H2SO4 and H^+ from HCN (weak acid).
You can go through the math but the second H from H2SO4 contributes only 0.0046 but compared to 0.036 from above probably should be added in. If you want to do that use
k2 = (H^+)(SO4^-)/(HSO4^-). For (H^+) substitute x+0.036 and for (SO4^2-) substitute x and for (HSO4^-) substitute 0.02-x. I get 0.00456M for k2 contribution from H2SO4. For the HCN; that is
HCN ==> H^+ + CN^-
Ka = (H^+)(CN^-)/(HCN) = 2.1E-9
(x+0.04056)/(0.018) = about 3E-6 which can be ignored.
Then pH = -log(H^+).
Post your work if you get stuck.
To calculate the pH of the given solution, we need to consider the acidic nature of the three compounds - HCl, H2SO4, and HCN.
1. Start by calculating the H+ ion concentration (also known as the hydronium ion concentration) for each compound using the given molarities:
- HCl: 1.6 x 10^-2 M
- H2SO4: 2.0 x 10^-2 M
- HCN: 1.8 x 10^-2 M
2. For a strong acid like HCl, all of it dissociates in water, so the H+ ion concentration is equal to the molarity of HCl: 1.6 x 10^-2 M.
3. H2SO4 is also a strong acid, so it also dissociates completely into H+ and SO4-2 ions. In this case, the concentration of H+ ions is also equal to the molarity of H2SO4: 2.0 x 10^-2 M.
4. HCN, however, is a weak acid and does not dissociate completely in water. To find the H+ ion concentration from a weak acid, we need to use the acid dissociation constant (Ka) for HCN.
5. The Ka value for HCN is 4.9 x 10^-10. To find the concentration of H+ ions, we can use an ICE (Initial, Change, Equilibrium) table and the equilibrium expression for the dissociation of HCN:
HCN (aq) ⇌ H+ (aq) + CN- (aq)
Let x be the concentration of H+ ions from the dissociation of HCN. At equilibrium, the concentration of HCN is (1.8 x 10^-2 - x) M, and the concentration of CN- ions is also x M.
Using the equilibrium expression: Ka = [H+][CN-] / [HCN]
Plug in the values:
4.9 x 10^-10 = x * x / (1.8 x 10^-2 - x)
Since the value of x from the dissociation is likely to be small compared to the initial concentration, we can approximate (1.8 x 10^-2 - x) as 1.8 x 10^-2.
Simplifying the equation gives:
x^2 = 4.9 x 10^-10 * 1.8 x 10^-2 = 8.82 x 10^-12
Taking the square root of both sides gives:
x = 9.38 x 10^-6 M (approx.)
Thus, the concentration of H+ ions from HCN is 9.38 x 10^-6 M.
6. Now, add up the H+ ion concentrations from all three compounds to get the total H+ ion concentration in the solution:
Total H+ ion concentration = concentration from HCl + concentration from H2SO4 + concentration from HCN
= 1.6 x 10^-2 M + 2.0 x 10^-2 M + 9.38 x 10^-6 M
= 3.6 x 10^-2 M (approx.)
7. Since the pH is defined as the negative logarithm (base 10) of the H+ ion concentration, we can calculate the pH using the formula:
pH = -log[H+]
Calculating the pH with the total H+ ion concentration:
pH = -log(3.6 x 10^-2)
≈ -log(3.6) + log(10^-2)
≈ -1.556 + 2
≈ 0.444
Therefore, the pH of the given solution is approximately 0.444.