Calculate the pH of (a) a solution that contains 0.10 M HNO3(aq) (Ka ~20 M) and 0.10 M HCN(aq) (Ka = 6.2 × 10–10 M), (b) a solution that contains 1.00 M HCN(aq) (Ka = 6.2 × 10–10 M) and 5.00 M HNO2(aq) (Ka = 4.0 × 10–4 M), and (c) 1.0 × 10–10 M HCl (aq). (Answer: (a) 1.00, (b) 1.35, (c) 7.00)
I know the answers to these problems but not really the process. I know how to individually solve for 0.10M HNO3 and 0.10M HCN but not together.
For (a) so you know HNO3 is a strong acid meaning that it ionizes 100% so the (H+) from HNO3 will be 0.1 M. You know HCN is a weak acid with a Ka of 6.2E-10. That tells you that the H+ contributed by the HCN is VERY small from HCN ==> H^+ + CN^-. Then to make it even smaller, the high concentration of H^+ from the HNO3 shifts that HCN equilibrium to the LEFT which makes the H^+ contribution from HCN EVEN SMALLER. So you ignore the H^+ from the HCN. So HNO3 contribution is 0.1 M and that gives you a pH of 1.00. BTW, what is that M by Ka? Two points. First, Ka has no units. Some may argue with that but technically that is correct since Ka is derived from activity coefficients and those have no units. Second, even if you scrape units out of a Ka (using formula units) I don't know of any that has units of M = molar. For b you have a solution of HNO2 and HCN. This is a weak acid (HNO2) and another weak acid (HCN). HNO2 is the stronger of the two (from the Ka values) PLUS it is five times stronger in molarity so again I would ignore the HCN.
...................HNO2 ==> H^+ + NO2^-
I....................5...............0..........0
C...................-x..............x...........x
E...................5-x..............x...........x
Ka = 4E-4 = (H^+)(NO2^-)/(HNO2)
4E-4 = (x)(x)/(5-x) Solve for x = about 0.044 (My calculator isn't working but that's about right and pH is about 1.35.
(c) is an old standby chemistry trick. Many students will give a pH of 10 BUT remember that is in H2O where the (H^+) = (OH^-) [without the HCl] so that becomes
...................HOH ==> H^+ + OH^-
I................................1E-7.........1E-7
add............................1E-10..........0
C.................................+1E-10......?
E.................................1E-7
In other words the H^+ from the HCl contributes so little to the 1E-7 already there that the final is essentially 1E-7 or pH = 7.00.
In mixtures like this you must think your way through them; i.e., ignore the one contributing so little. Remember, too, that you may have a mixture of a weak acid/salt or weak base/salt and these you tackle with the Henderson-Hasselbalch equation. Hope this helps.
To solve the problems, we will use the concept of the ionization constant (Ka) and the expression for the pH of a solution.
The ionization constant (Ka) is a measure of how much a weak acid or base dissociates in water. It is defined as the ratio of the concentration of the products of the dissociation to the concentration of the undissociated species.
The expression for the pH of a solution is given by:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution.
Let's solve each part step-by-step:
(a) Solution with 0.10 M HNO3(aq) and 0.10 M HCN(aq):
1. Write the chemical equation for the dissociation of each weak acid:
HNO3(aq) → H+(aq) + NO3-(aq)
HCN(aq) → H+(aq) + CN-(aq)
2. Calculate the concentration of hydrogen ions (H+) from the dissociation of each weak acid:
[H+] = [HNO3] = 0.10 M (since HNO3 fully dissociates)
[H+] = [HCN] = √(Ka * [HCN]) = √((6.2 × 10^-10 M) * (0.10 M)) ≈ 7.8 × 10^-6 M
3. Calculate the pH using the expression pH = -log[H+]:
pH = -log(7.8 × 10^-6) ≈ 5.11
(b) Solution with 1.00 M HCN(aq) and 5.00 M HNO2(aq):
1. Write the chemical equation for the dissociation of each weak acid:
HCN(aq) → H+(aq) + CN-(aq)
HNO2(aq) → H+(aq) + NO2-(aq)
2. Calculate the concentration of hydrogen ions (H+) from the dissociation of each weak acid:
[H+] = [HCN] = √(Ka * [HCN]) = √((6.2 × 10^-10 M) * (1.00 M)) ≈ 7.9 × 10^-6 M
[H+] = [HNO2] = √(Ka * [HNO2]) = √((4.0 × 10^-4 M) * (5.00 M)) ≈ 0.02 M
3. Calculate the pH using the expression pH = -log[H+]:
pH = -log(7.9 × 10^-6 + 0.02) ≈ 1.35
(c) Solution with 1.0 × 10^-10 M HCl(aq):
1. Write the chemical equation for the dissociation of HCl:
HCl(aq) → H+(aq) + Cl-(aq)
2. Since HCl is a strong acid, it fully dissociates, and the concentration of H+ is equal to the initial concentration:
[H+] = [HCl] = 1.0 × 10^-10 M
3. Calculate the pH using the expression pH = -log[H+]:
pH = -log(1.0 × 10^-10) ≈ 10
Please let me know if you need any further assistance.
To calculate the pH of a solution that contains multiple acids, you need to consider the individual dissociation constants (Ka) and the concentrations of each acid.
(a) Solution with 0.10 M HNO3(aq) and 0.10 M HCN(aq):
First, let's calculate the pH due to HNO3.
HNO3 is a strong acid, so we can assume it dissociates completely in water.
HNO3 → H+ + NO3-
Since the concentration of HNO3 is 0.10 M, the concentration of H+ is also 0.10 M. Thus, the pH due to HNO3 is -log(0.10) = 1.00.
Now, let's consider the contribution of HCN to the pH.
HCN → H+ + CN-
The dissociation constant (Ka) of HCN is 6.2 × 10–10 M.
To calculate the concentration of H+ from the dissociation of HCN, we need to consider the equilibrium expression for Ka:
Ka = [H+][CN-] / [HCN]
Since the concentration of HCN is 0.10 M and the concentration of CN- is negligible, we can assume that the concentration of HCN decreases by x and the concentration of H+ increases by x. Thus, we can set up an equilibrium expression:
6.2 × 10–10 = x * x / (0.10 - x)
Since x is small compared to 0.10, we can approximate that [HCN] ≈ 0.10 M. Therefore, the equation becomes:
6.2 × 10–10 = x * x / 0.10
Solving this equation, we find that x is approximately 7.87 × 10–6 M.
Now, we can calculate the concentration of H+ from HCN, which is about 7.87 × 10–6 M. Taking the negative logarithm of this concentration, we get the pH due to HCN, which is approximately -log(7.87 × 10–6) ≈ 5.05.
To find the overall pH of the solution, we sum up the contributions from HNO3 and HCN:
pH = 1.00 + 5.05 = 6.05
Rounded to two decimal places, the pH is approximately 6.05.
(b) Solution with 1.00 M HCN(aq) and 5.00 M HNO2(aq):
Follow a similar process as in part (a) but substitute the concentrations and Ka values of HCN and HNO2. You should find that the overall pH is approximately 1.35.
(c) Solution with 1.0 × 10–10 M HCl(aq):
Since HCl is a strong acid, it dissociates completely in water. The concentration of H+ is therefore 1.0 × 10–10 M. Taking the negative logarithm of this concentration gives us the pH, which is approximately -log(1.0 × 10–10) = 10.
However, note that this solution is quite dilute, so the activity coefficient of H+ should be considered to get a more accurate pH. Assuming an activity coefficient of approximately 0.92, the pH would be slightly lower, around 7.00.
Thus, the answers are: (a) 6.05, (b) 1.35, and (c) 7.00.