The solubility of highly soluble, thermally unstable salts such as ammonium chloride may be determined by back titration. In one experiment a 5.00 mL saturated solution of ammonium chloride, NH4Cl, at 20.0 °C, was diluted with distilled water to 250.0 mL in a standard flask. 20.0 mL aliquot of this solution was added to 10.0 mL of 0.400 M NaOH solution. The solution was heated to drive off the ammonia formed by this reaction. When the flask had cooled, the excess hydroxide ions were neutralised by 14.7 mL of 0.125 M HCl solution. The molar mass of ammonium chloride is 53.5 g mol−1. i. Write an equation for the neutralisation reaction. ii. Determine the amount, in mole, of NaOH that was originally added to the ammonium chloride solution. iii. Determine the amount, in mole, of ammonium chloride in the 20.0 mL aliquot. iv. Calculate the amount, in mole, of ammonium chloride in 5.00 mL of the saturated solution. v. Calculate the solubility, in gL−1, of ammonium chloride in water at 20 °C. b. If the burette was rinsed with water instead of acid before the titration, how would the calculated solubility of ammonium chloride be affected? Explain your answer.

SSCE VCE Chemistry Measuring Solubility and Concentration: The solubility of highly soluble

The solubility of highly soluble, thermally unstable salts such as ammonium chloride may be determined by back titration - VCE - SSCE Chemistry - Question 8 - 2012 - Paper 1

Question 8

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The solubility of highly soluble, thermally unstable salts such as ammonium chloride may be determined by back titration. In one experiment a 5.00 mL saturated solu... show full transcript

Worked Solution & Example Answer:The solubility of highly soluble, thermally unstable salts such as ammonium chloride may be determined by back titration - VCE - SSCE Chemistry - Question 8 - 2012 - Paper 1

Step 1

i. Write an equation for the neutralisation reaction.

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Answer

The neutralisation reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) with ammonium chloride (NH4Cl) can be represented by the following equation:

$\text{NaOH} (aq) + \text{HCl} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2\text{O} (l)$

Step 2

ii. Determine the amount, in mole, of NaOH that was originally added to the ammonium chloride solution.

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Answer

To find the amount of NaOH, we can use the concentration and volume of the NaOH solution:

$n(\text{NaOH}) = C \times V = 0.400 \, \text{mol L}^{-1} \times 0.0100 \, \text{L} = 0.00400 \, \text{mol}$

Step 3

iii. Determine the amount, in mole, of ammonium chloride in the 20.0 mL aliquot.

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Answer

We first need to determine the amount of excess NaOH that was neutralised by HCl:

Using the neutralisation volume of HCl:

$n(\text{HCl}) = C \times V = 0.125 \, \text{mol L}^{-1} \times 0.0147 \, \text{L} = 0.00184 \, \text{mol}$

Since the reaction between NaOH and HCl is a 1:1 ratio, the amount of NaOH that was neutralised is also 0.00184 mol.

Thus, the amount of NaOH that reacted with NH4Cl is:

$n(\text{NaOH}) \text{(initial)} - n(\text{NaOH}) \text{(excess)} = 0.00400 \, \text{mol} - 0.00184 \, \text{mol} = 0.00216 \, \text{mol}$

Step 4

iv. Calculate the amount, in mole, of ammonium chloride in 5.00 mL of the saturated solution.

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Answer

The saturated solution was diluted to 250 mL. Thus, to find the number of moles of ammonium chloride in the entire solution, apply the dilution factor:

$n(\text{NH}_4\text{Cl}) \text{ in } 250 \, \text{mL} = 0.00216 \, \text{mol} \times \frac{250 \, \text{mL}}{20.0 \, \text{mL}} = 0.0270 \, \text{mol}$

Now, to find the concentration in 5 mL:

$n(\text{NH}_4\text{Cl}) = 0.0270 \, ext{mol} \times \frac{5.0 \, \text{mL}}{250 \, \text{mL}} = 0.00054 \, \text{mol}$

Step 5

v. Calculate the solubility, in gL−1, of ammonium chloride in water at 20 °C.

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Answer

To find the solubility, consider the total amount in 1 L of solution (1000 mL):

$\text{Solubility (g L}^{-1}\text{)} = \left(\frac{n(\text{NH}_4\text{Cl})}{1 \, \text{L}}\right) \times \text{Molar Mass} = 0.0270 \, \text{mol} \times 53.5 \, \text{g mol}^{-1} = 1.44 \, ext{g L}^{-1}$

Step 6

If the burette was rinsed with water instead of acid before the titration, how would the calculated solubility of ammonium chloride be affected? Explain your answer.

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Answer

If the burette was rinsed with water, the volume of HCl added would be greater than necessary, leading to a higher amount of acid neutralising the NaOH. This increases the calculated amount of NaOH consumed and hence the amount of NH4Cl determined. Consequently, the calculated solubility of ammonium chloride would be inaccurately reported as higher than the actual value.

The solubility of highly soluble, thermally unstable salts such as ammonium chloride may be determined by back titration - VCE - SSCE Chemistry - Question 8 - 2012 - Paper 1

Worked Solution & Example Answer:The solubility of highly soluble, thermally unstable salts such as ammonium chloride may be determined by back titration - VCE - SSCE Chemistry - Question 8 - 2012 - Paper 1

i. Write an equation for the neutralisation reaction.

ii. Determine the amount, in mole, of NaOH that was originally added to the ammonium chloride solution.

iii. Determine the amount, in mole, of ammonium chloride in the 20.0 mL aliquot.

iv. Calculate the amount, in mole, of ammonium chloride in 5.00 mL of the saturated solution.

v. Calculate the solubility, in gL−1, of ammonium chloride in water at 20 °C.

If the burette was rinsed with water instead of acid before the titration, how would the calculated solubility of ammonium chloride be affected? Explain your answer.

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