向10mL0.01mol/LBaCl2溶液中加入10mL0.01mol/LH2SO4溶液,反应后溶液的ph为
问题描述:
向10mL0.01mol/LBaCl2溶液中加入10mL0.01mol/LH2SO4溶液,反应后溶液的ph为
答
c(H+) = 2n(H2SO4)/V(混合后的体积) = 2*0.01*0.01/0.02 = 0.01 mol/L
PH = -lg c(H+) = -lg0.01 = 2n是指硫酸物质的量么?对啊,n(H2SO4)=c(H2SO4)*V(H2SO4)