Question

Calculate the [H+] and [OH-] of each of the following solutions. 0.010 M HCl 0.010 M H2SeO4 0.025 M KOH 0.020 M Ba(OH)2

Answer 1

0.010 M HCl

We know that HCl is a strong acid. Hence, it dissociates according to the equation,

HCl -----------> H+ + Cl-

From the balanced equation, it is clear that, 1 M HCl produces 1M of H+ and 1 M of Cl-.

Hence, [H+] = [Cl-] = [HCl]

So, [H+] = 0.010 M

We know that, [H+] [OH-] = 1 x 10^-14

So, [OH-] = (1 x 10^-14) / 0.010 = 1 x 10^-12

[OH-] = 1 x 10^-12

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0.010 M H2SeO4

We know that H2SeO4 is a strong acid. Hence, it dissociates according to the equation,

H2SeO4 -----------> 2H+ + SeO4^2-

From the balanced equation, it is clear that, 1 M H2SeO4 produces 2M of H+ and 1 M of SeO4^2-.

Hence, [H+] = 2 x 0.010 M = 0.020 M

[H+] = 0.020 M

We know that, [H+] [OH-] = 1 x 10^-14

So, [OH-] = (1 x 10^-14) / 0.020 = 5 x 10^-13

[OH-] = 5 x 10^-13

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0.025 M KOH

We know that KOH is a strong base. Hence, it dissociates according to the equation,

KOH -----------> K+ + OH-

From the balanced equation, it is clear that, 1 M KOH produces 1M of K+ and 1 M of OH-.

Hence, [K+] = [OH-] = [KOH]

So, [OH-] = 0.025 M

We know that, [H+] [OH-] = 1 x 10^-14

So, [H+] = (1 x 10^-14) / 0.025 = 4 x 10^-13

[OH-] = 4 x 10^-13

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0.020 M Ba(OH)2

We know that Ba(OH)2 is a strong base. Hence, it dissociates according to the equation,

Ba(OH)2 -----------> Ba+ + 2OH-

From the balanced equation, it is clear that, 1 M Ba(OH)2 produces 1M of Ba+ and 2 M of OH-.

Hence, [Ba+] = 2[OH-] = [Ba(OH)2]

So, [OH-] = 2 x 0.020 M = 0.040 M

[OH-] = 0.040 M

We know that, [H+] [OH-] = 1 x 10^-14

So, [H+] = (1 x 10^-14) / 0.040 = 2.5 x 10^-13

[H+] = 2.5 x 10^-13