Question

What volume of a $0.50$ $M$ solution of hydrochloric acid is required to neutralize $60.0$ millilitres of a $1.50$ $M$ potassium hydroxide solution?

• A. $360$ ml
• B. $180$ ml
• C. $120$ ml
• D. $60.0$ ml
• E. $20.0$ ml

Answer 1

Answer: (B)

$180$ ml

$HCl+KOH\to KCl+H_{2}O$
$MV=M^{\prime }{V}^{\prime }$
$M$ is molarity of $HCl$ (0.50 M)
$V$ is volume of $HCl$.
$M^{\prime }$ is molarity of $KOH$ (1.50 M).
$V^{\prime }$ is volume of $KOH$ (60.0 ml)
$0.500V=1.50\times 60.0$
$V=180$ ml
Hence, 180 ml of a $0.50M$ solution of hydrochloric acid is required to neutralize 60.0 milliliters of a $1.50M$ potassium hydroxide solution.