Câu 3:
a.
$n HCl$=$\frac{0,73}{36,5}$ =0,02 (mol)
$n H_2SO_4$=$\frac{0,98}{98}$ =0,01 (mol)
⇒$∑n [H^+]$=$n HCl$+$2n H_2SO_4$=0,02+0,01.2=0,04 (mol)
⇒$[H^+]$=$\frac{0,04}{2}$ =0,02 (mol)
⇒pH=-log$[H^+]$=(-log(0,02)≈1,7
b.
$n KOH$=$\frac{2,8}{56}$ =0,05 (mol)
$n Ba(OH)_2$=$\frac{4,275}{171}$ =0,025 (mol)
⇒$∑n [OH^-]$=$n KOH$+$2n Ba(OH)_2$=0,05+0,025.2=0,1 (mol)
⇒$[OH^-]$=$\frac{0,1}{5}$ =0,02 (mol)
⇒pH=14+log$[OH^-]$=14+log(0,02)≈12,3
--------------------Nguyễn Hoạt------------------
