Question
The initial rate of reaction was determined for the reaction between bromide ions, \(Br^−\)(aq), and bromate ions, \(BrO_3^−\) (aq), in acidic solution.
\(5Br^− (aq) + BrO_3^− (aq) + 6H^+ (aq) → 3Br_2 (aq) + 3H_2O(l)\)
Four trials were carried out and the results are given in the following table.
(a) Deduce the overall rate equation.
(b) The rate of a reaction at 50°C is three times the rate at 25°C. Calculate the activation energy, in kJ \(mol^{−1}\) , for this reaction using sections 1 and 2 of the data booklet.
Answer/Explanation
Answer:
(a) \(H^+\): second order
AND
\(BrO_3^−\): first order
AND
\(Br^−\): first order
«rate =» k[\(Br^−\)][\(BrO_3^−][H^+]^2\)
(b) 
Question
Methanoic acid is a monoprotic weak acid.
(a) The concentration of methanoic acid was found by titration with a 0.200mol \(dm^{−3}\) standard solution of sodium hydroxide, NaOH(aq), using an indicator to determine the end point.
(i) Calculate the pH of the sodium hydroxide solution.
(ii) Write an equation for the reaction of methanoic acid with sodium hydroxide.
(iii) 22.5\(cm^3\) of NaOH(aq) neutralized 25.0\(cm^3\) of methanoic acid. Determine the concentration of the methanoic acid.
(iv) Calculate the pH of the original solution of methanoic acid. Use your answer to (a)(iii) and section 21 of the data booklet. If you did not get an answer to (a)(iii) use 0.300mol \(dm^{−3}\), but this is not the correct answer.
(v) Identify, giving a reason, a suitable indicator for this titration. Use section 22 of the data booklet.
(b) Write an ionic equation to show why a solution of sodium methanoate does not have a pH of 7.
Answer/Explanation
Answer:
(a) (i) «[\(OH^−\)] = 0.200 mol \(dm^{−3}\)»
ALTERNATIVE 1:
«pOH = −\(log_{10}\)(0.200) =» 0.699 ü
«pH = 14.000 – 0.699 =» 13.301
ALTERNATIVE 2:
«[\(H^+\)] = \(\frac{1.00 \times 10^{-14}}{0.200}\) = » 5.00 \(\times 10^{-14}\) «mol \(dm^{−3}\)»
«pH = −\(log_{10}\)(5.00 x \(10^{−14}\))» = 13.301
(ii) HCOOH(aq) + NaOH(aq) \(\rightarrow \) HCOONa(aq) + \(H_2O(l)\)
(iii) «n(acid) = n(\(OH^−\))»
«[acid] = \(\fraction {0.200 mol dm^{-3} \times 22.5 \times 10^{-3} dm^3}{25.0 \times 10^{-3} dm^3}\) » = 0.180 «mol \(dm^{−3}\)»
(iv) 
(v) phenolphthalein
OR
phenol red
colour change in the pH range of equivalence point
(b) \(HCOO^−\)(aq) + \(H_2O(l)\) \(\rightarrow \) HCOOH(aq) + \(OH^−\)(aq)