Question

Calculate the [H3O+] of each aqueous solution with the following [OH−]:

Part A

NaOH, 1.6×10⁻² M ,

milk of magnesia, 1.2×10⁻⁵ M,

aspirin, 2.2×10⁻¹¹ M

seawater, 3.5×10⁻⁶ M

Express your answer using two significant figures.

Answer 1

Sol :-

(a). For NaOH :-

Given [OH^-] = 1.6×10⁻² M

We know that

K_w = [H₃O⁺] [OH^-] , here K_w is ionic product of water whose value = 1.0 x 10^-14 at 25 ⁰C

also

[H₃O⁺] = K_w /  [OH^-]

[H₃O⁺] = 1.0 x 10^-14 /  [OH^-] ..............(1)

[H₃O⁺] = 1.0 x 10^-14 / 1.6×10⁻² M

[H₃O⁺] = 6.2 x 10^-13 M

(b). For Milk of magnesia :-

Given [OH^-] = 1.2×10⁻⁵ M

From equation (1), we have

[H₃O⁺] = 1.0 x 10^-14 / 1.2×10⁻⁵ M

[H₃O⁺] = 8.3 x 10^-10 M

(c). For aspirin : -

Given [OH^-] = 2.2×10⁻¹¹ M

From equation (1), we have

[H₃O⁺] = 1.0 x 10^-14 / 2.2×10⁻¹¹ M

[H₃O⁺] = 4.5 x 10^-4 M

(d). For seawater : -

Given [OH^-] = 3.5×10⁻⁶ M

From equation (1), we have

[H₃O⁺] = 1.0 x 10^-14 / 3.5×10⁻⁶ M

[H₃O⁺] = 2.9 x 10^-9 M