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AQA Jun 2017 Paper 1 Q2
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![0 2 0 2 . 1This question is about acidic solutions. The acid dissociation constant, Ka, for ethanoic acid is given by the expression Ka = [CH COO 3 [CH 3 [H ] COOH] The value of Ka for ethanoic acid is 1.74105 mol dm3 at 25 C A buffer solution with a pH of 3.87 was prepared using ethanoic acid and sodium ethanoate. In the buffer solution, the concentration of ethanoate ions was 0.136 mol dm3 Calculate the concentration of the ethanoic acid in the buffer solution. Give your answer to three significant figures. [3 marks] Concentration of acid mol dm3 *04* IB/M/Jun17/7405/1<br />
0 2 . 2 *05*In a different buffer solution, the concentration of ethanoic acid was 0.260 mol dm3 and the concentration of ethanoate ions was 0.121 mol dm3 A 7.00103 mol sample of sodium hydroxide was added to 500 cm3 of this buffer solution. Calculate the pH of the buffer solution after the sodium hydroxide was added. Give your answer to two decimal places. [6 marks] pH of buffer solution Turn overIB/M/Jun17/7405/1<br />](https://learnah.org/wp-content/uploads/2018/08/June_2017_QP_-_Paper_1_AQA_Chemistry_A-Level_2-0.jpg)
![MARK SCHEMEA-LEVEL CHEMISTRY7405/1JUNE 2017 Question Answers Mark Additional Comments/Guidance [H+] = (103.87 =) 1.3489 x 104 [CH3COOH] = [] = ( [+][] [.][.] [.] = 1.05436) 1.051.06 (mol dm3) 02.1 Allow 1.35 x 10-4. If M1 wrong can only score M2. Mark is for correctly rearranged equation. 3 sf or more 16 of 35<br />
If 0.007 moles in 500 cm3 seen follow Mark Scheme 1 Mark Scheme 1 moles ethanoic acid = 0.130 moles sodium ethanoate = 0.0605 mol CH3COOH after addition = (0.130 - 0.007 ) = 0.123 mol CH3COO- after addition = (0.0605+0.007) = 0.0675 [H+] =( [ ][] [] ) = (= 3.171 x 10 -5 ) [.][.] [.] pH = 4.50 (must be 2dp) If 0.014 moles in 1 dm3 follow Mark Scheme 2 Mark scheme 2 02.2 moles CH3COOH after addition = (0.260 - 0.014 ) = 0.246 (This scores 2 marks) moles CH3COO- after addition = (0.121+0.014) = 0.135 (This scores 2 marks) [H+] =( [ ][] [] ) = [.][.] [.] pH = 4.50 (must be 2dp) MARK SCHEMEA-LEVEL CHEMISTRY7405/1JUNE 2017 1 1 1 1 1 1 Method 1 For M3 allow M10.007 For M4 allow M2 + 0.007 Method 1 and 2 M5 = expression with their numbers M6 = answer to 2 dp pH = 4.50 scores 6 marks Ifused in Ka expression, stop at M4 If divide by 2 after M5, lose M6 Allow solutions which use Henderson-Hasselbach Equation 17 of 35<br />](https://learnah.org/wp-content/uploads/2018/08/June_2017_MS_-_Paper_1_AQA_Chemistry_A-Level_2-0.jpg)
