Photo AI

The amount of pure HCl gas, in mol, that must be added to the solution to lower the pH from 12.0 to 2.0 would be A - VCE - SSCE Chemistry - Question 16 - 2007 - Paper 1

Question icon

Question 16

The-amount-of-pure-HCl-gas,-in-mol,-that-must-be-added-to-the-solution-to-lower-the-pH-from-12.0-to-2.0-would-be-A-VCE-SSCE Chemistry-Question 16-2007-Paper 1.png

The amount of pure HCl gas, in mol, that must be added to the solution to lower the pH from 12.0 to 2.0 would be A. 10 B. 2.0 C. 0.02 D. 0.01

Worked Solution & Example Answer:The amount of pure HCl gas, in mol, that must be added to the solution to lower the pH from 12.0 to 2.0 would be A - VCE - SSCE Chemistry - Question 16 - 2007 - Paper 1

Step 1

Calculate the initial concentration of H⁺ ions at pH 12.0

96%

114 rated

Answer

To find the concentration of H⁺ ions at pH 12.0, we can use the formula:

[H+]=10pH[H^+] = 10^{-pH}

Thus, [H+]=1012.0=1.0×1012 mol/L[H^+] = 10^{-12.0} = 1.0 \times 10^{-12} \text{ mol/L}

Step 2

Calculate the final concentration of H⁺ ions at pH 2.0

99%

104 rated

Answer

For pH 2.0, the concentration of H⁺ ions is:

[H+]=102.0=0.01 mol/L[H^+] = 10^{-2.0} = 0.01 \text{ mol/L}

Step 3

Determine the change in concentration

96%

101 rated

Answer

The change in concentration of H⁺ ions needed:

Δ[H+]=[H+]final[H+]initial=0.011.0×10120.01 mol/L\Delta [H^+] = [H^+]_{final} - [H^+]_{initial} = 0.01 - 1.0 \times 10^{-12} \approx 0.01 \text{ mol/L}

Step 4

Calculate the volume of the solution

98%

120 rated

Answer

Assuming we are adding HCl to 1 liter of solution, the amount needed will be equal to the change in concentration:

Amount of HCl needed=0.01extmoles\text{Amount of HCl needed} = 0.01 ext{ moles}

Step 5

Final Answer

97%

117 rated

Answer

Hence, the amount of pure HCl gas that must be added is approximately 0.01 mol, so the answer is D. 0.01.

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;