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19 Nov 2019
Might need to zoom in on computer, but I canât because I had to fit the whole question
please fill out table and answer with this given information:
with the information above, answer this please
This is just a review of what we learned about the common ion effect. Remember, if there is any weak acid or weak base present, use that equilibrium reaction as your equation. In this case, we have HCOOH, so it's dissociation reaction will be our equation. As we have discussed, the concentration of the [H30+ is not actually equal to zero at the start of the reaction, it is just a convenient (while still accurate) way to solve the problem HCOOH (aq) H2o (I) HCOO (aq) H30 (ag) 0.930 M t x 0.930+x 1.04 M + X 1.04-x Bccause the simplifying assumption is valid here we get Ka 6.80 x 104 - [HCOO30.930x) [HCOOH] Therefore: [H3O+] = x = 7.60 x 10-4 And the initial pH of the buffer is set at pH 3.12
Might need to zoom in on computer, but I canât because I had to fit the whole question
please fill out table and answer with this given information:
with the information above, answer this please
This is just a review of what we learned about the common ion effect. Remember, if there is any weak acid or weak base present, use that equilibrium reaction as your equation. In this case, we have HCOOH, so it's dissociation reaction will be our equation. As we have discussed, the concentration of the [H30+ is not actually equal to zero at the start of the reaction, it is just a convenient (while still accurate) way to solve the problem HCOOH (aq) H2o (I) HCOO (aq) H30 (ag) 0.930 M t x 0.930+x 1.04 M + X 1.04-x Bccause the simplifying assumption is valid here we get Ka 6.80 x 104 - [HCOO30.930x) [HCOOH] Therefore: [H3O+] = x = 7.60 x 10-4 And the initial pH of the buffer is set at pH 3.12
Nestor RutherfordLv2
21 Feb 2019