Calculate the concentration of the acid solution (HY) based on the titration results and given information.

Question

ScanSolve AI Tutor · Accepted Answer

To solve this problem, we must first make an assumption for the average volume of acid A (HY) used in the titration, as the experimental procedure in part (a) was not performed.

Assumption for part (a):
We assume that 25.00 cm³ of NaOH solution (B) was titrated with acid HY (A).
Let the average volume of acid A (HY) used from the burette be $\text{24.50 cm}^3$ .

Given information:

Mass of HY dissolved to make 250 cm³ solution = 1.64 g
Concentration of NaOH solution (B) = 0.100 mol dm⁻³
Volume of NaOH solution (B) used = 25.00 cm³
Reaction: $\text{HY}_{(\text{aq})} + \text{NaOH}_{(\text{aq})} \longrightarrow \text{NaY}_{(\text{aq})} + \text{H}_2\text{O}_{(\text{l})}$
Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0

b) From your results and the information provided, calculate the:

Calculate the moles of NaOH used.
The concentration of NaOH is given, and the volume used is the pipette volume.
$\text{Moles of NaOH} = \text{Concentration of NaOH} \times \text{Volume of NaOH}$ $n_{\text{NaOH}} = 0.100 \text{ mol dm}^{-3} \times \left(\frac{25.00}{1000}\right) \text{ dm}^3$ $n_{\text{NaOH}} = 0.100 \text{ mol dm}^{-3} \times 0.02500 \text{ dm}^3 = 0.002500 \text{ mol}$
Determine the moles of HY that reacted.
From the balanced chemical equation, the mole ratio of HY to NaOH is 1:1.
$n_{\text{HY}} = n_{\text{NaOH}} = 0.002500 \text{ mol}$
Calculate the concentration of A (HY) in mol dm⁻³.
This is the concentration of the acid in the burette, using the assumed average titre volume.
$\text{Concentration of HY} = \frac{\text{Moles of HY}}{\text{Volume of HY used}}$ $C_{\text{HY}} = \frac{0.002500 \text{ mol}}{\left(\frac{24.50}{1000}\right) \text{ dm}^3}$ $C_{\text{HY}} = \frac{0.002500 \text{ mol}}{0.02450 \text{ dm}^3} = 0.10204 \text{ mol dm}^{-3}$
Rounding to three significant figures:
$\text{Concentration of A (HY)} = \boxed{\text{0.102 mol dm}^{-3}}$
Calculate the concentration of A (HY) in g dm⁻³.
The acid HY was prepared by dissolving 1.64 g in 250 cm³ of solution.
$\text{Concentration of HY in g dm}^{-3} = \frac{\text{Mass of HY}}{\text{Volume of solution}}$ $\text{Concentration of HY in g dm}^{-3} = \frac{1.64 \text{ g}}{\left(\frac{250}{1000}\right) \text{ dm}^3}$ $\text{Concentration of HY in g dm}^{-3} = \frac{1.64 \text{ g}}{0.250 \text{ dm}^3} = 6.56 \text{ g dm}^{-3}$ $\text{Concentration of A (HY)} = \boxed{\text{6.56 g dm}^{-3}}$
Calculate the molar mass of the acid HY.
The molar mass can be found by dividing the concentration in g dm⁻³ by the concentration in mol dm⁻³.
$\text{Molar mass of HY} = \frac{\text{Concentration of HY in g dm}^{-3}}{\text{Concentration of HY in mol dm}^{-3}}$ $M_{\text{HY}} = \frac{6.56 \text{ g dm}^{-3}}{0.10204 \text{ mol dm}^{-3}} = 64.288 \text{ g mol}^{-1}$
Rounding to three significant figures:
$\text{Molar mass of HY} = \boxed{\text{64.3 g mol}^{-1}}$
*c

Calculate the concentration of the acid solution (HY) based on the titration results and given information.

Answer

Step-by-step solution

Calculate the moles of NaOH used.

Determine the moles of HY that reacted.

Calculate the concentration of A (HY) in mol dm⁻³.

Calculate the concentration of A (HY) in g dm⁻³.

Calculate the molar mass of the acid HY.

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