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December 18, 2014

December 18, 2014

Posted by **Tom** on Tuesday, May 27, 2008 at 9:36pm.

Solution A is 50 mL of 0.1 M solution of the weak monoprotic acid HX.

Solution B is a 0.05 M solution of the salt NaX. It has a pH of 10.02.

Solution C is made by adding 15 mL of 0.25 M KOH to Solution A.

What are you suppose to do with Solution B to help solve for solution A.

My work:

10^-10.02 = 9.55x 10^-11= [H+]

[OH-]= 1.047 x 10^.4

Is 9.55 x 10^-11 the Ka of Solution B.

NaX--Na+ + X

X + H2O-- HX + OH

wat should i do to solve for B to solve for A.

what should I do after this

Responses

Chemistry - DrBob222, Tuesday, May 27, 2008 at 9:02pm

Use solution B to solve for the Ka of the weak acid.

Use Ka from above to solve for the pH of solution A. That gives you the first 3 numbers of the combination.

Use solution C and the Henderson-Hasselbalch equation to solve for the pH of that solution which will be last three numbers of the combination.

Chemistry - Tom, Tuesday, May 27, 2008 at 9:36pm

how do u use Solution B to solve for the Ka?

- chemistry -
**DrBob222**, Tuesday, May 27, 2008 at 10:09pmSolution B is 0.05 M solution of NaX.

The hydrolysis of X^- is

X^- + HOH ==> HX + OH^-

Kb = Kw/Ka = (HX)(OH^-)/(X^-)

You know pH is 10.02. Subtract from 14 to get pOH and solve for (OH^-) remembering that pOH= -log(OH^-). You know (HX) = (OH^-). YOu know Kw and (X^-) so the only unknown is Ka. Solve for Ka.

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