Enough of a monoprotic acid is dissolved in water to produce a 0.0131 M solution. The pH of the resulting solution is 2.46. Calculate the Ka for the acid.
I calculated [H^+] to be 3.46727 x 10^-3.. Then I used this 'x' value and plugged it into Ka = (x)^2 / (0.0119-x).... I got Ka to equal 1.25 x 10^-3 but it's wrong... Help please? Thank you!!
I calculated [H^+] to be 3.46727 x 10^-3.. Then I used this 'x' value and plugged it into Ka = (x)^2 / (0.0119-x).... I got Ka to equal 1.25 x 10^-3 but it's wrong... Help please? Thank you!!
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Where is the .0119 coming from? Shouldn't this be .0131? At first I thought it was just a typo, but I don't get 1.25 x 10^-3 with either number. I get 9.2 x 10^-4.
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[H+] = 10^-2.46 = 3.46737 x 10^-3 M (not 3.46727)
Ka = [(3.46737 x 10^-3) (3.46737 x 10^-3)] / 0.0131
no need for 0.0131 minus x
Ka = 9.2 x 10^-4
Ka = [(3.46737 x 10^-3) (3.46737 x 10^-3)] / 0.0131
no need for 0.0131 minus x
Ka = 9.2 x 10^-4