Tuesday
March 28, 2017

Post a New Question

Posted by on .

If 10.0 mL of 0.250 mol/L NaOH(aq) is added to 30.0 mL of 0.17 mol/L HOCN(aq), what is the pH of the resulting solution?

  • chemistry - ,

    NaOH + HOCN ==> NaOCN + HOH
    mols NaOH to begin = M x L = 0.250 x 0.010 = ??

    mols HOCN to begin = M x L = 0.17 x 0.030 = ??

    There is more HOCN than NaOH. The difference in mols is how much HOCN is left unreacted. The NaOCN formed is the amount of the lesser chemical (in this case NaOH). So you have a buffer formed consisting of a weak acid (HOCN) and its salt (NaOCN).
    Use the Henderson-Hasselbalch equation.
    pH = pKa + log [(base)/(acid)]
    Post your work if you get stuck.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question