Posted by **vivian ** on Thursday, May 12, 2011 at 9:54pm.

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

- chemistry -
**DrBob222**, Thursday, May 12, 2011 at 10:21pm
...............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.

- chemistry -
**vivian **, Thursday, May 12, 2011 at 10:40pm
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

- chemistry -
**vivian **, Thursday, May 12, 2011 at 10:41pm
The problem didn't include Ka

- chemistry -
**DrBob222**, Thursday, May 12, 2011 at 11:14pm
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.

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