温度为25℃时,将0.23g钠投入到100ml水中充分反应后所得溶液的密度为1g/cm3,则该溶液的PH为( ) A.1 B.13 C.12 D.10
问题描述:
温度为25℃时,将0.23g钠投入到100ml水中充分反应后所得溶液的密度为1g/cm3,则该溶液的PH为( )
A. 1
B. 13
C. 12
D. 10
答
令生成氢气的质量为m,NaOH的物质的量为n,则:
2Na+2H2O=2NaOH+H2↑
46g 2mol 2g
0.23g n m
m=
=0.01g0.23g×2g 46g
n=
=0.01mol0.23mol×2mol 46g
故溶液质量=0.23g+100mL×1g/mL-0.01g=100.22g
故溶液体积=
=100.22mL100.22g 1g/mL
故c(OH-)=c(NaOH)=
=0.1mol/L,0.01mol 0.10022L
故c(H+)=
mol/L=10-13mol/L,10−14
0.1
故pH=-lg10-13=13,
故选B.