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Calculate the pH of the following solutions at 298 K.
a) 0.25 mol dm-3 HCl.​​​​​​​​
b) 0.15 mol dm-3 H2SO4.​​​
c) 0.15 mol dm-3 benzoic acid. Ka(benzoic acid) = 6.4 x 10-5 mol dm-3.
d) 0.10 mol dm-3 KOH. Kw(298 K) = 1.0 x 10-14 mol2 dm-6.

Calculate The PH Of The Following Solutions At 298 K A 025 Mol Dm3 HCl B 015 Mol Dm3 H2SO4 C 015 Mol Dm3 Benzoic Acid Kabenzoic Acid 64 X 105 Mol Dm3 D 010 Mol class=

Sagot :

pstnonsonjoku

The term pH has to do with the hydrogen ion concentration of a solution.

What is pH?

The term pH has to do with the hydrogen ion concentration of a solution. Let us now proceed the pH of the solutions;

a) pH of HCl = -log(0.25) = 0.62

b) pH of H2SO4 = -log(0.15) = 0.82

c)  We have to set up an ICE table here;

              C6H5COOH + H2O ⇄ C6H5COO- + H^+

I            0.15                                  0                      0

C          -x                                       +x                     +x

E         0.15 -x                                  x                      x

Ka = [C6H5COO-] [ H^+]/[C6H5COOH]

6.4 x 10^-5 = x^2/0.15 - x  

6.4 x 10^-5(0.15 - x) = x^2

9.6 * 10^-6 - 6.4 x 10^-5x = x^2

x^2 + 6.4 x 10^-5x - 9.6 * 10^-6 = 0

x=0.003 M

Hence;

[C6H5COO-] = [ H^+] = x = 0.003 M

pH = - log[0.003 M] = 2.5

d) [OH-] [H^+] = 1.0 x 10-14 mol2 dm-6

[H^+] = 1.0 x 10-14 mol2 dm-6/ 0.10 mol dm-3

[H^+] = 1 * 10^-13 M

pH= - log[1 * 10^-13 M] = 13

Learn more about pH of a solution:https://brainly.com/question/15289741

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