Question

1 263mL的NH4I溶液,[NH4I]=0.300M,NH3的Kb=1.74*10^(-5),求NH4I溶液的pH值

Answer 1

NH4I 的物质的量浓度 c = 0.300 mol/L,

[H+] = √ (cKa) = √ (cKw/Kb) = √ [0.3*10^(-14) /(1.74*10^(-5)) ] = 1.31*10^(-5) mol/L,

所以pH = -lg [H+] = 4.88.

n(NaOH) = n(HBr) = 0.0620 * 65 mmol.

V(NaOH) = 0.0620 * 65 / 0.035 mL = 115.14 mL