Question

you weigh out 0.110g hypobromous acid into a 250ml volumetric flask. if you have to titrate this solution against 0.11m naoh, how much naoh would you need?
a 0.01ml
b 0.10l
c 10.3ml
d 100.3ml
continue
question 4 of 6

Explanation:

Step1: Calculate the moles of hypobromous acid ($\ce{HBrO}$)

The molar mass of $\ce{HBrO}$ is $M=(1 + 79.904+16)\space g/mol = 96.904\space g/mol$.
Using the formula $n=\frac{m}{M}$, where $m = 0.110\space g$ and $M = 96.904\space g/mol$.
So, $n=\frac{0.110\space g}{96.904\space g/mol}\approx1.135\times 10^{-3}\space mol$.

Step2: Determine the mole ratio from the reaction equation

The reaction between $\ce{HBrO}$ and $\ce{NaOH}$ is $\ce{HBrO + NaOH
ightarrow NaBrO + H2O}$.
The mole ratio of $\ce{HBrO}$ to $\ce{NaOH}$ is $1:1$. So, $n(\ce{NaOH})=n(\ce{HBrO}) = 1.135\times 10^{-3}\space mol$.

Step3: Calculate the volume of $\ce{NaOH}$

Using the formula $V=\frac{n}{C}$, where $n = 1.135\times 10^{-3}\space mol$ and $C = 0.11\space M$.
$V=\frac{1.135\times 10^{-3}\space mol}{0.11\space mol/L}\approx0.0103\space L$.
Convert liters to milliliters: $V = 0.0103\space L\times1000\space mL/L = 10.3\space mL$.

Answer:

C. 10.3 mL