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Chemistry

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Q3. A third beaker, C, contains 100.0 cm^3 of 0.0125 mol/dm^3 ethanoic acid ( Ka = 1.74 × 10^−5 mol/dm^3 at 25 ºC). Write an expression for Ka and use it to calculate the pH of the ethanoic acid solution in beaker C.

A3. Ethanoic acid is a weak acid with the formula CH3COOH. It will dissociate into H+ and CH3COO- therefore the expression for Ka is [H+][CH3COO-]/[CH3COOH] The [H+] = [CH3COO-] and so the expression ...

Answered by Tutor51285 D. • Chemistry tutor

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Q2. Calculate the pH of the solution formed after 50.0 cm^3 of 0.0108 mol/dm^3 aqueous sodium hydroxide are added to beaker B. Give your answer to 2 decimal places

A2. First thing to work out is the moles of NaOH added, since moles = concentration x volume, = 0.0108 mol/dm^3 x 50ml, but need volume to be in units of Litres to match with the units of dm^3 so the fo...

Answered by Tutor51285 D. • Chemistry tutor

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Q1. Two beakers, A and B, each contain 100.0 cm^3 of 0.0125 mol/dm^3 nitric acid. Calculate the pH of the solution formed after 50.0 cm^3 of distilled water are added to beaker A. Give your answer to 2 decimal places.

A1. First it is important to know that nitric acid (HNO3) is a strong acid and so [HNO3] = [H+] Beaker A contains 100cm^3 = 100ml Final volume is 150ml (50cm^3 of distilled water is added to beaker...

Answered by Tutor51285 D. • Chemistry tutor

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balance the following equation: Na2O + HCl --> NaCl + H2O

Na₂O + 2HCl --> 2NaCl + H₂O, easiest way to do this is to balance this equation is to write down all the atoms on both sides of the reaction. Its easiest to start with an element ...

Answered by Stephen A. • Chemistry tutor

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