求溶液的PH值问题?

一块表面有氧化钠的金属钠5.4g,放入1000ml水中,放入1.12升H2(标况),将此溶液稀释为2升,此溶液的PH值?

热心网友

2Na + 2 H2O === 2 NaOH + H2Na2O + H20 === 2 NaOH nH2 = 1.12/22.4 =0.05 mol所以,nNa =0.05 *2 = 0.1mol mNa = 0.1 *23 =2.3gmNa2O = 5.4-2.3= 3.1g n Na2O = 3.1/62 = 0.05 moln NaOH =nNa + 2 nNa2O = 0.1 + 0.05*2 =0.2 molnOH- =0.2/2 =0.1mol/LnH+ = 1*10^-14 /0.1 =1*10^-13 mol/L所以：PH= 13

热心网友

思路：楼上太罗嗦了。可想而知最后变成NaOH，OH-和Na+物质量相等，只要算Na的物质量就行。设有Na2O 物质量x2Na + 2 H2O === 2 NaOH + H2nH2 = 1.12/22.4 =0.05 mol所以,nNa =0.05 *2 = 0.1mol5.4=0.1*23+62x ；x=0.05mol;所以有Na+物质量0.1+0.05*2=0.2mol;下面你自己来，加油