High School Chemistry HW

  • Thread starter jacker2011
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  • #1

Homework Statement

Find pH of solution A and that of Solution C with the following information below:

Solution A:50ml of 0.100 M solution of weak monoprotic acid HX
Solution B: 0.0500M solution of salt NaX. It has a pH of 10.02
Solution C: made by adding 15.0m; of 0.250 M KOH to solution A

Homework Equations

Ka for monoprotic acid is 7.2 x 10^-4, Ka=[H+] [anion]. Anion= x

The Attempt at a Solution

i try to find [H+] which i did was 1/2log(Ka)=
1/2log (7.2 x 10^-4) = 0.0268
or I find the M of [H+] in solution A
and i found 0.005M of HX

Answers and Replies

  • #2
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The simplification not being immediately obvious to me, I find that the equilibrium constant expression for the weak acid could generally be:
[tex] \[
K_a = \frac{{[H][H]}}{{F_a - [H]}}
\] [/tex]
Which is easily transformed to [tex] \[
[H]^2 + K_a [H] - K_a F = 0
\] [/tex]
You can from there use the solution to a quadratic equation. In that equation, [tex]\[F_a \] [/tex] is formality of the acid, and [H] is molarity of hydrogen ion (or hydronium).

For solution #c, note if you have excess HX or excess KOH. If you have excess HX, then you have a solution containing HX and KX, meaning this is a buffer. Base your calculations on the modification required for the Ka formula (my best guess, right now - depends on relative ratios and strength of Ka and Kb; the KX has a Kb value while the HX has a Ka value.)
  • #3
What jacker2011 aims at is [H+]=sqrt(Ka*C), which works for dissociation degree smaller than 5% (see discussion of calculation of pH of weak acid for details).

Buffer solution is not about containing HX and KX, but about containing HX and X- at the same time (ie acid and conjugated base). That's close, but it is not the same.