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Calculate the pH of a 0.2 mol L⁻¹ solution of hydrochloric acid - HSC - SSCE Chemistry - Question 17 - 2006 - Paper 1

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Calculate the pH of a 0.2 mol L⁻¹ solution of hydrochloric acid. Calculate the pH after 20 mL of 0.01 mol L⁻¹ sodium hydroxide is added to 50 mL of 0.2 mol L⁻¹ hydr... show full transcript

Worked Solution & Example Answer:Calculate the pH of a 0.2 mol L⁻¹ solution of hydrochloric acid - HSC - SSCE Chemistry - Question 17 - 2006 - Paper 1

Step 1

Calculate the pH of a 0.2 mol L⁻¹ solution of hydrochloric acid.

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Answer

To calculate the pH of a hydrochloric acid (HCl) solution, we use the formula:

pH=log[H+]\text{pH} = -\log[\text{H}^+]

Since hydrochloric acid is a strong acid, it fully dissociates in solution:

HClH++Cl\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-

For a 0.2 mol L⁻¹ HCl solution, the concentration of hydrogen ions [H+][\text{H}^+] is also 0.2 mol L⁻¹.

Now, substituting into the pH formula:

pH=log(0.2)0.70\text{pH} = -\log(0.2) \approx 0.70

Step 2

Calculate the pH after 20 mL of 0.01 mol L⁻¹ sodium hydroxide is added to 50 mL of 0.2 mol L⁻¹ hydrochloric acid.

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Answer

First, we calculate the moles of HCl and NaOH:

  1. Moles of HCl:

Moles=Concentration×Volume=0.2 mol L1×0.050 L=0.01 mol\text{Moles} = \text{Concentration} \times \text{Volume} = 0.2 \text{ mol L}^{-1} \times 0.050 \text{ L} = 0.01 \text{ mol}

  1. Moles of NaOH:

Moles=0.01 mol L1×0.020 L=0.0002 mol\text{Moles} = 0.01 \text{ mol L}^{-1} \times 0.020 \text{ L} = 0.0002 \text{ mol}

Next, we write the balanced chemical equation:

HCl+NaOHNaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

Now we subtract the moles of NaOH from moles of HCl:

Remaining moles of HCl=0.01 mol0.0002 mol=0.0098 mol\text{Remaining moles of HCl} = 0.01 \text{ mol} - 0.0002 \text{ mol} = 0.0098 \text{ mol}

The final total volume after mixing is:

Total Volume=50 mL+20 mL=70 mL=0.070 L\text{Total Volume} = 50 \text{ mL} + 20 \text{ mL} = 70 \text{ mL} = 0.070 \text{ L}

Now, calculate the new concentration of HCl:

[H+]=0.0098 mol0.070 L0.14 mol L1[\text{H}^+] = \frac{0.0098 \text{ mol}}{0.070 \text{ L}} \approx 0.14 \text{ mol L}^{-1}

Finally, calculate the pH:

pH=log(0.14)0.85\text{pH} = -\log(0.14) \approx 0.85

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