What is the pH of 8.94 × 10−6 M HCN(aq), ignoring the effect of the autoprotolysis of water?
1. 7.17923 (basic; thus not reasonable)
2. 6.03594
3. 7.30027 (basic; thus not reasonable)
4. 5.50032
5. 7.38101 (basic; thus not reasonable)
6. 5.18207
7. None of these
...............HCN ==>H^+ + CN^-
iniyisl....8.94E-6.....0.....0
change.......-x.........x.....x
equil....8.94E-6 - x.....x....x
Ka = (H^+)(CN^-)/(HCN)
Look up Ka, substitute from the above ICE chart and solve the quadratic equation for x, then convert to pH.
When you post problems like this you should include the Ka values for the numbers we look up in our texts may not agree with the ones in your text. Using my value for Ka the answer is about 6.9.
i did it and used 6.2E-10. my answer wasn't there so i was wondering if i did something wrong..i got 7.11 for pH
The problem didn't include Ka
Using your value of 6.2E-10 for Ka for HCN, I solved the quadratic and obtained 7.41E-8 for a pH of 7.13. Obviously this can't be right for 7.13 is basic which means that one cannot neglect the ionization of water. You may want to check your solution of the quadratic; I've gone over mine several times and I don't find an error. If you confirm that number I would go with "none of these" for an answer. I'm surprised that one of the answer is not 1,3, or 5.
To find the pH of 8.94 × 10^−6 M HCN(aq), we need to use the expression for the acidic dissociation of HCN.
The dissociation of HCN can be represented by the equation:
HCN(aq) ⇌ H^+(aq) + CN^-(aq)
The dissociation constant for HCN is given by:
K_a = [H^+(aq)][CN^-(aq)] / [HCN(aq)]
Since the concentration of water is much higher compared to the concentration of HCN, we can ignore the autoprotolysis of water and assume that [H2O] remains constant.
The concentration of [HCN(aq)] is given as 8.94 × 10^−6 M.
Let's assume x is the concentration of H^+ ions formed when HCN dissociates. Thus, the concentration of CN^- ions formed when HCN dissociates is also x.
Using the expression for the dissociation constant:
K_a = x * x / (8.94 × 10^−6)
Simplifying the equation:
K_a = x^2 / (8.94 × 10^−6)
Rearranging the equation:
x^2 = K_a * (8.94 × 10^−6)
Taking the square root of both sides:
x = √(K_a * 8.94 × 10^−6)
Next, we need to find the value of the dissociation constant, K_a. The pK_a value of HCN is given as 9.34. The pK_a is defined as -log(K_a), so we can find K_a using the expression:
K_a = 10^(-pK_a)
Substituting the value of pK_a, we get:
K_a = 10^(-9.34)
Now, substitute the value of K_a back into the equation for x:
x = √(10^(-9.34) * 8.94 × 10^−6)
After calculating the value of x, take the negative logarithm to find the pH value:
pH = -log(x)
Once you have calculated the value of pH, compare it with the given options to find the correct answer.