![Calculate either [H,O*] or [OH-] for each of the solutions. Solution A: OH) = 2.71 x 10-PM Solution A: [H,0) = Solution B: [H](http://img.homeworklib.com/questions/0e09b5b0-35d2-11eb-828c-210db0ddc33b.png?x-oss-process=image/resize,w_560)
Homework Answers
Using formula, for the autoionization of water
Kw = [H3O^+][OH^-]
where Kw = water dissociation constant
[H3O^+] => concentration of acid in solution
[OH^-] => concentration of base in solution
At 25°C, Kw => 1.0 * 10^-14 M^2
Solution A : [OH^-] = 2.71 * 10^-7 M
Thus, [H3O^+] = ( 1 * 10^-14) / ( 2.71 * 10^-7) M
=> [H3O^+] = 0.369 * 10^-7 or 3.69 * 10^-8 M
Hence Solution A, [H3O^+] => 3.69 * 10^-8 M
Now, pH => -log[H3O^+] => -log(3.69*10^-8) => 7.43 (approx)
Since pH> 7, Hence, Solution A is basic.
Solution B : [H3O^+] = 7.63 * 10^-9 M
Thus, [OH^-] = ( 1 * 10^-14) / ( 7.63 * 10^-9) M
=> [OH^-] = 0.131 * 10^-5 M => 1.31 * 10^-6 M
Hence Solution B, [OH^-] => 1.31 * 10^-6 M
Now, pH = -log[H3O^+] => -log(7.63*10^-9) => 8.12 (approx)
Since, pH>7, Hence, solution B is basic.
Solution C: [H3O^+] = 6.79 * 10^-4 M
Thus, [OH^-] = ( 1 * 10^-14) / ( 6.79 * 10^-4) M
=> [OH^-] = 0.147 * 10^-10 M => 1.47 * 10^-11 M
Hence Solution C, [OH^-] => 1.47 * 10^-11 M
Now, pH = -log[H3O^+] => -log(6.79 * 10^-4) => 3.17(approx)
Since, pH<7, Hence, solution C is acidic.
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