 # Question & Answer: Write the equilibrium constant expression for the following reaction. CaCO_3(s) +….. Write the equilibrium constant expression for the following reaction. CaCO_3(s) + H_2SO_4(aq) rightarrow CaSO_4(aq) + CO_2(g) + H_2O(1) At 0 degree C, K_w is about 1.14 times 10^-15, what is the pH under these conditions? Will an aqueous solution of NH_3 be acidic, neutral or basic? Why? (I am looking for how the substance dissolves and comparing the products to the definitions of acid, base, salt.)

a)

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solid and liquid are not considered while writing Kc expression

SO,

Kc = [CaSO4][CO2]/[H2SO4]

b)

Kw = 1.14*10^-15

use:

Kw = [H+] [OH-]

since water is neutral, [H+] = [OH-]

Kw = [H+] [H+]

Kw = [H+]^2

1.14*10^-15 = [H+]^2

[H+]= 3.376*10^-8

pH = -log[H+]

pH = -log(3.376*10^-8)

pH = 7.47