Question

if a 25.0 ml sample of sulfuric acid is titrated with 50.0 ml of 0.200 M potassium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?

A) 0.150 M
B) 0.100 M
C) 0.200 M
D) 0.300 M
E) 0.400 M

Answer 1

Answer:

Molarity of the sulfuric acid is Choice C): 0.200 M.

Explanation:

The unit for molarity "M" stands for moles per liter. Convert all volume to liters:

• $V({\rm NaOH}) = \rm 50.0\;ml = 50.0\times 10^{-3}\;L = 5.00\times 10^{-2}\;L$;
• $V({\rm H_2SO_4}) = \rm 25.0\;ml = 25.0\times 10^{-3}\;L = 2.50\times 10^{-2}\;L$.

How many moles of potassium hydroxide is required to reach the equivalence endpoint?

$n({\rm NaOH}) = c \cdot V = 0.200\;\text{mol}\cdot\text{L}^{-1} \times 5.00times 10^{-2} \;\text{L} = 1.00\times 10^{-2}\;\text{mol}$.

How many moles of sulfuric acid in the original solution?

$\rm KOH$ neutralizes $\rm H_2SO_4$ at a two-to-one ratio:

$\rm 2\;KOH\;(aq) +H_2SO_4\;(aq) \to K_2SO_4\;(aq) + H_2O\;(l)$.

The $1.00\times 10^{-2}\;\text{mol}$ moles of $\rm KOH$ will neutralize only half as much $\rm H_2SO_4$. That is:

$\displaystyle n({\rm H_2SO_4}) = \rm \frac{1}{2}\times (1.00\times 10^{-2}\;mol)= 5.00\times 10^{-3}\;mol$.

What's the molarity of the $\rm H_2SO_4$ solution?

$\displaystyle \begin{aligned}M({\rm H_2SO_4}) &= \frac{n(\rm H_2SO_4)}{V({\rm H_2SO_4})}=\rm \frac{5.00\times 10^{-3}\;mol}{2.50\times 10^{-2}\;L} = 0.200\;mol\cdot\L^{-1} = 0.200\;M\end{aligned}$.