Question

欲配制500mlpH=9 氨根离子浓度为1mol/L的缓冲溶液，需密度为0.904g/mol、氨的质量分数为26%的浓氨水多少克

最好有讲解做题思路

Answer 1

(1) PH ==9
POH == 14-9 == 5
[OH-] == 10^-5 mol/L
(2) 1.8^-5 == 10^-5 *C(NH3)/C(NH4+)
1.8 ==C(NH3)/C(NH4+)
C(NH4+) == 1mol/L
C(NH3) == 1.8mol/L
(3)m(NH4CI) == 53.5 *1*0.5 == 26.75g
(4) 1.8*0.5 == 0.904v*26% /17
v == 65.1ml
(5)称量26.8克氯化铵，量取65.1毫升氨水配成500毫升溶液即为所求！