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Question: 25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volum...

25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solution?

A

25 mL

B

50 mL

C

12.5 mL

D

75 mL

Answer

25 mL

Explanation

Solution

The problem involves two acid-base titrations.

First titration: 25 mL of HCl solution reacts with 30 mL of 0.1 M sodium carbonate (Na2CO3\text{Na}_2\text{CO}_3) solution. The reaction between HCl and Na2CO3\text{Na}_2\text{CO}_3 is: 2HCl+Na2CO3→2NaCl+H2O+CO22\text{HCl} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 From the stoichiometry, 2 moles of HCl react with 1 mole of Na2CO3\text{Na}_2\text{CO}_3.

We can use the concept of millimoles. Number of millimoles of Na2CO3=Molarity×Volume (in mL)\text{Na}_2\text{CO}_3 = \text{Molarity} \times \text{Volume (in mL)} Millimoles of Na2CO3=0.1 mmol/mL×30 mL=3 mmol\text{Na}_2\text{CO}_3 = 0.1 \text{ mmol/mL} \times 30 \text{ mL} = 3 \text{ mmol}

According to the stoichiometry, the number of millimoles of HCl reacted is twice the number of millimoles of Na2CO3\text{Na}_2\text{CO}_3. Millimoles of HCl = 2×Millimoles of Na2CO3=2×3 mmol=6 mmol2 \times \text{Millimoles of } \text{Na}_2\text{CO}_3 = 2 \times 3 \text{ mmol} = 6 \text{ mmol}

These 6 millimoles of HCl are present in 25 mL of the HCl solution. The molarity of the HCl solution (MHClM_{\text{HCl}}) is: MHCl=Millimoles of HCl/Volume of HCl (in mL)M_{\text{HCl}} = \text{Millimoles of HCl} / \text{Volume of HCl (in mL)} MHCl=6 mmol/25 mL=0.24 mmol/mL=0.24 MM_{\text{HCl}} = 6 \text{ mmol} / 25 \text{ mL} = 0.24 \text{ mmol/mL} = 0.24 \text{ M}

Second titration: The same HCl solution is used to titrate 30 mL of 0.2 M aqueous NaOH solution. The reaction between HCl and NaOH is: HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} From the stoichiometry, 1 mole of HCl reacts with 1 mole of NaOH.

We are given: Volume of NaOH solution (VNaOHV_{\text{NaOH}}) = 30 mL Molarity of NaOH solution (MNaOHM_{\text{NaOH}}) = 0.2 M

Number of millimoles of NaOH = Molarity×Volume (in mL)\text{Molarity} \times \text{Volume (in mL)} Millimoles of NaOH = 0.2 mmol/mL×30 mL=6 mmol0.2 \text{ mmol/mL} \times 30 \text{ mL} = 6 \text{ mmol}

According to the stoichiometry, the number of millimoles of HCl required is equal to the number of millimoles of NaOH. Millimoles of HCl required = Millimoles of NaOH = 6 mmol

Let VHCl2V_{\text{HCl2}} be the volume of the HCl solution required in mL. We know the molarity of the HCl solution is 0.24 M. Millimoles of HCl required = MHCl×VHCl2M_{\text{HCl}} \times V_{\text{HCl2}} 6 mmol=0.24 mmol/mL×VHCl26 \text{ mmol} = 0.24 \text{ mmol/mL} \times V_{\text{HCl2}} VHCl2=6/0.24 mLV_{\text{HCl2}} = 6 / 0.24 \text{ mL} VHCl2=600/24 mLV_{\text{HCl2}} = 600 / 24 \text{ mL} VHCl2=25 mLV_{\text{HCl2}} = 25 \text{ mL}