一塊表面有氧化鈉的金屬鈉5.4g,放入1000ml水中,放入1.12升H2(標況),將此溶液稀釋為2升,此溶液的PH值?

熱心網友

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

熱心網友

思路：樓上太羅嗦了?？上攵詈笞兂蒒aOH，OH-和Na+物質量相等，只要算Na的物質量就行。設有Na2O 物質量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+物質量0.1+0.05*2=0.2mol;下面你自己來，加油