# Question #88405

Aug 5, 2017

$0.01 m o l$ of $N a O H$

#### Explanation:

Reaction:

$C {H}_{3} C O O H + N a O H \to C {H}_{3} C O O N a + {H}_{2} O$

$M = 0.1 M$

$V = 0.1 L$

$M = \frac{n}{V}$

$n = 0.01 m o l$

$1 m o l$ of $C {H}_{3} C O O H$ requires $1 m o l$ of $N a O H$

$\therefore 0.01 m o l$ of $C {H}_{3} C O O H$ requires $1 \times 0.1 m o l$ of $N a O H$

$= 0.01 m o l$ of $N a O H$