Question

1: calculate the titer of a 0.01135 mol·l⁻¹ hcl solution against naoh. mnaoh=40.01 g.mol⁻¹

Explanation:

Step1: Write the reaction equation

$HCl + NaOH=NaCl + H_2O$
The mole - ratio of $HCl$ to $NaOH$ is $n_{HCl}:n_{NaOH}=1:1$.

Step2: Define the titer formula

The titer of $HCl$ against $NaOH$ ($T_{HCl/NaOH}$) is defined as the mass of $NaOH$ (in grams) that reacts with 1 mL of the $HCl$ solution.
We know that $n = cV$. For a volume $V = 1\ mL=1\times10^{- 3}\ L$ of the $HCl$ solution with concentration $c_{HCl}=0.01135\ mol\cdot L^{-1}$, the number of moles of $HCl$ in 1 mL of the solution is $n_{HCl}=c_{HCl}\times V$.
$n_{HCl}=0.01135\ mol\cdot L^{-1}\times1\times10^{-3}\ L = 1.135\times10^{-5}\ mol$
Since $n_{HCl}=n_{NaOH}$, and $m = nM$ (where $m$ is mass, $n$ is number of moles and $M$ is molar mass), and $M_{NaOH}=40.01\ g\cdot mol^{-1}$.

Step3: Calculate the mass of $NaOH$

$m_{NaOH}=n_{NaOH}\times M_{NaOH}$
Substitute $n_{NaOH}=1.135\times10^{-5}\ mol$ and $M_{NaOH}=40.01\ g\cdot mol^{-1}$ into the formula.
$m_{NaOH}=1.135\times10^{-5}\ mol\times40.01\ g\cdot mol^{-1}=4.541\times10^{-4}\ g$

Answer:

$4.541\times10^{-4}\ g/mL$