Question

:Trộn lẫn V ml dung dịch NaOH 0,01M với V ml dung dịch HCl 0,03M, thu được 2V ml dung dịch Y. Dung dịch Y có pH bằng

Answer 1

\(n_{OH^-}=0.01\cdot10^{-3}\cdot V\left(mol\right)\)

\(n_{H^+}=0.03\cdot10^{-3}\cdot V\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

\(n_{H^+\left(dư\right)}=0.03\cdot10^{-3}V-0.01\cdot10^{-3}V=0.02\cdot10^{-3}V\left(mol\right)\)

\(\left[H^+\right]=\dfrac{0.02\cdot10^{-3}}{2}=0.01\cdot10^{-3}=10^{-5}\)

\(pH=-log\left(10^{-5}\right)=5\)

Answer 2

$n_{NaOH} = 0,001V.0,01(mol) < n_{HCl} = 0,001V.0,03(mol)$

Do đó axit dư

$NaOH + HCl \to NaCl + H_2O$

$n_{H^+\ dư} = n_{HCl\ dư} = 0,001V.0,03 - 0,001V.0,01 = 0,001V.0,02(mol)$
$[H^+] = \dfrac{0,001V.0,02}{0,002V} = 0,01$
$pH = -log(0,01) = 2$