# Question 80880

Mar 19, 2015

There were 0.04 mol of HA in 100 mL of solution.

You didn't have to do a titration, since you knew the concentration of the original solution.

You started with 110 mL of 0.5 mol/L HA and diluted it to a total of 150mL.

To calculate the concentration of the dilute solution, you can use the dilution formula

${c}_{1} {V}_{1} = {c}_{2} {V}_{2}$, where $c$ is the concentration and $V$ is the volume

${c}_{1} = \text{0.5 mol/L}$; ${V}_{1} = \text{110 mL}$
${c}_{2} = \text{?}$; ${V}_{2} = \text{150 mL}$

c_2 = c_1 × V_1/V_2 = "0.5 mol/L" × (110 cancel("mL"))/(150 cancel("mL")) = "0.37 mol/L"

The moles of HA in 100 mL of this solution are

0.1000 cancel("L soln") × "0.37 mol HA"/(1 cancel("L soln")) = "0.04 mol HA"

Note: The answer can have only 1 significant figure, because that is all you gave for the concentration of the HA. If you need more precision, you will have to recalculate.

The reaction for the titration was

HA + NaOH → NaA + H₂O

The moles of HA in the sample you titrated were

0.03430 cancel("L NaOH") × (0.1 cancel("mol NaOH"))/(1 cancel("L NaOH")) × "1mol HA"/(1 cancel("mol NaOH")) = "0.0034 mol HA"

But this was for 10.00 mL of solution. In 100 mL of solution there would be

$\text{0.0034 mol HA" × (100 cancel("mL"))/(10.00 cancel("mL")) = "0.03 mol HA}$

Note: The answer can have only 1 significant figure, because that is all you gave for the concentration of the NaOH. If you need more precision, you will have to recalculate.

Mar 19, 2015

The answer is indeed $\text{0.04 moles}$ of acetic acid in 100.0-mL of the original solution.

I would only like to point out that you don't actually need to perform a dilution calculation and figure out the concentration of the 150.0-mL acetic acid solution, you can work directly with moles.

So, you know that you had 110.0 mL of 0.5 M acetic acid. You figure out the number of moles of acetic acid present in this sample

$C = \frac{n}{V} \implies {n}_{\text{acetic}} = C \cdot V$

n_("acetic") = "0.5 M" * 110.0 * 10^(-3)"L" = "0.055 moles"#

The acetic acid solution, the one with the added 40.0 mL of water, will have 0.055 moles of acetic acid, because adding water will not affect the number of moles of acetic acid already present, it will only affect the solution's concentration.

You can use a simple proportion to figure out how many moles of acetic acid you have in a 100-mL sample of this 150.0-mL solution

$\text{100 mL solution" * "0.055 moles HA"/"150 mL solution" = "0.0367 moles HA}$

Like Ernest said, the answer can have no more than one sig fig, and no titration was actually needed to determine what you had to determine.