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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.)
Expert Answer
a)
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
Answer: 7.47
c)
NH3 has lone pair on Nitrogen. Due to this it acts as base
NH3 dissolves in water as:
NH3 + H2O —> NH4+(aq) + OH- (aq)
Since solution has OH- ions
It is basic
