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7.1 Two acids, HX and HY, of EQUAL CONCENTRATIONS are compared - NSC Physical Sciences - Question 7 - 2022 - Paper 2

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7.1 Two acids, HX and HY, of EQUAL CONCENTRATIONS are compared. The pH of HX is 2.7 and the pH of HY is 0.7. 7.1.1 Define an acid in terms of the Lowry-Brønsted the... show full transcript

Worked Solution & Example Answer:7.1 Two acids, HX and HY, of EQUAL CONCENTRATIONS are compared - NSC Physical Sciences - Question 7 - 2022 - Paper 2

Step 1

7.1.1 Define an acid in terms of the Lowry-Brønsted theory.

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Answer

According to the Lowry-Brønsted theory, an acid is defined as a proton (H⁺) donor. This means that an acid is a substance that can donate a hydrogen ion to another substance during a chemical reaction.

Step 2

7.1.2 Which acid, HX or HY, is STRONGER? Give a reason for the answer.

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Answer

HX is the stronger acid because it has a higher pH (2.7) compared to HY, which has a pH of 0.7. A lower pH indicates a higher concentration of hydronium ions (H₃O⁺), meaning HY is more ionized and thus a stronger acid.

Step 3

7.1.3 Is the concentration of the hydronium ions HIGHER THAN, LOWER THAN or EQUAL TO the concentration of HX₂? Give a reason for the answer.

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Answer

The concentration of the hydronium ions is LOWER THAN the concentration of HX₂. This is due to the fact that HX has a relatively low Kₐ value of 1.8 × 10⁻⁵, indicating that it does not ionize completely in solution, resulting in fewer hydronium ions being produced.

Step 4

7.2.1 Concentration of the H₃O⁺ ions in the final solution

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Answer

The pH of the final solution is given as 2. Therefore, the concentration of hydronium ions (H₃O⁺) can be calculated using the formula:

[H3O+]=10extpH[H₃O^+] = 10^{- ext{pH}}

Substituting in the pH value:

[H3O+]=102=0.01extmoldm3[H₃O^+] = 10^{-2} = 0.01 ext{ mol·dm}^{-3}

Step 5

7.2.2 Concentration of the NaOH(aq)

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Answer

Apply the stoichiometry of the reaction. Given that the concentration of HCl is 0.03 mol·dm⁻³ and its volume is 0.2 dm³, the moles of HCl used are:

n(HCl)=cimesV=0.03extmoldm3imes0.2extdm3=0.006extmoln(HCl) = c imes V = 0.03 ext{ mol·dm}^{-3} imes 0.2 ext{ dm}^3 = 0.006 ext{ mol}

Since the reaction between HCl and NaOH is 1:1, the moles of NaOH will be equal to the moles of HCl:

n(NaOH)=0.006extmoln(NaOH) = 0.006 ext{ mol}

Now, using the volume of NaOH, we can find its concentration. The volume of NaOH is 150 cm³ or 0.15 dm³:

c(NaOH) = rac{n(NaOH)}{V(NaOH)} = rac{0.006 ext{ mol}}{0.15 ext{ dm}^3} = 0.04 ext{ mol·dm}^{-3}

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