Answer to Question #141914 in General Chemistry for Gabriella Santorelli

Question

What mass of NaOH should be added to 500.0ml of 0.015M H₂SO₄ to produce a solution with a pH of 2.00? (assume no volume change)

Answer 1

$\begin{aligned} 2NaOH + H_2SO_4\quad^{\longrightarrow} _{\longleftarrow}\quad Na_2SO_4 + 2H^+ + 2OH^- \end{aligned}$

$\begin{aligned} \textsf{Amount of } H_2SO_4 &= 0.015M × 0.5L \\ &= 0.0075mol \end{aligned}$

$\begin{aligned} \therefore \textsf{Amount of }[H^+] \textsf{ reacting in the reaction}&= 2(0.0075mol)\\ &= 0.015mol \end{aligned}$

$\begin{aligned} pH &= -\log[H^+]\\ \textsf{Concentration of unreacted }[H^+] &= 10^{-pH}\\ &= 10^{-2.00}\\ &=0.01M \end{aligned}$

$\textsf{Since there is no volume change, we can assume }$$NaOH \textsf{ does not add to the total volume.}$

$\begin{aligned} \therefore \textsf{Amount of }H^+ \textsf{ left unreacted }&= 0.01M × 0.5L\\ &= 0.005mol \end{aligned}$

From the table above, it is shown how the amount of $OH^-$ is 0.010mol

From the chemical reaction above,

2 moles of $2OH^-$ is produced by 2 moles of $NaOH$

$\therefore 0.010$ moles of $2OH^-$ is produced by $0.010$ moles of$NaOH$

$\begin{aligned} \textsf{Mass of }NaOH &= 0.010moles × 40g/mol\\ &= 0.4g \end{aligned}$