acids,bases and buffers Flashcards
What is meant by a weak acid and Brønsted–Lowry acid
Proton/H+ donor
AND
Partially dissociates/ionises
Calculate the pH of 0.500 mol dm–3 potassium hydroxide.
H+] = 1.00 ×10–14 0.5(00) OR 2(.00) × 10–14 (mol dm–3) pH = –log 2(.00) × 10–14 = 13.7(0)
Write the full equation for the reaction of aqueous propanoic acid with sodium carbonate.
2C2H5COOH + Na2CO3 → 2C2H5COONa + CO2 + H2O
Write the ionic equation for the reaction of aqueous propanoic acid with aqueous potassium hydroxide.
H+ + OH– → H2O
A small amount of aqueous ammonia, NH3(aq), is added to the buffer solution.Explain, in terms of equilibrium, how the buffer solution would respond to the added NH3(aq).
Added ammonia C2H5COOH removes added NH3/alkali/base OR C2H5COOH + NH3 / OH– → OR NH3/alkali reacts with/accepts H+ OR H+ + NH3 → OR H+ + OH– → Equlibrium → C2H5COO– OR Equilibrium → right
How can an aqueous solution of an acid contain hydroxide ions?
Water dissociates/ionises OR H2O ⇌ H \+ \+ OH – OR 2H2O ⇌ H3O \+ \+ OH–
Write a full equation for the reaction between ethanoic acid and solid calcium carbonate.
2CH3COOH + CaCO3 → (CH3COO)2Ca + CO2 + H2O
Explain why this buffer solution has formed.
solution contains CH3COOH AND CH3COO–
Explain how this buffer solution controls pH when either an acid or an alkali is added.
Quality of written communication, QWC
2 marks are available for explaining how the equilibrium
system allows the buffer solution to control the pH on addition
of H+
and OH-
(see below)
——————————————————
CH3COOH ⇌ H
+
+ CH3COO–
—————————————————–
CH3COOH reacts with added alkali
OR CH3COOH + OH–
OR added alkali reacts with H+
OR H
+
+ OH–
Equilibrium right OR Equilibrium CH3COO– (QWC)
CH3COO–
reacts with added acid
Equilibrium left OR Equilibrium CH3COOH (QWC)
0.14 mol dm–3 solutions of hydrochloric acid, HCl, and chloric(I) acid, HClO (pKa = 7.43), have different pH values.Explain why the pH values are different and calculate the pH of 0.14 mol dm–3 solutions of HCl and HClO to two decimal places.
Show any working in calculations.
HCl is a strong acid AND HClO is a weak acid HCl: pH = –log 0.14 = 0.85 (2 DP required) HClO: CHECK THE ANSWER ON ANSWER LINE IF answer = 4.14, award all three calculation marks -------------------------------------------------------------------- Ka = 10–7.43 OR 3.7 x 10–8 (mol dm–3) [H+] = Ka [HClO] OR Ka [HA] OR Ka 0.14 OR 3.7 x 108 x 0.14 pH = 4.14 (2 DP required)
Aluminium powder is added to aqueous ethanoic acid, CH3COOH.Write full and ionic equations for the reaction that takes place.
2Al + 6CH3COOH 2(CH3COO)3Al + 3H2
2Al + 6H+ 2Al3+ + 3H2
Calculate the pH of a 0.40 mol dm–3 solution of NaOH.
[H+] = Kw [OH ] OR 1.0 1014 [OH ] OR 1.0 1014 0.4(0) OR 2.5 x 10–14 (mol dm–3) Correctly calculates pH = –log 2.5 x 10–14 = 13.6(0)
Explain what is meant by the term buffer solution.
Describe how a buffer solution based on methanoic acid can act as a buffer.
A buffer solution minimises pH changes on addition of small amounts of acid/H+ or alkali/OH–/base -------------------------------------------------------------------------------- HCOOH ⇌ H+ + HCOO– Equilibrium sign essential Added alkali HCOOH reacts with added alkali/base/OH– OR added alkali/OH– reacts with H+ QWC: Equilibrium shifts forming HCOO– OR H+ OR (HCOOH) Equilibrium right Added acid HCOO– reacts with added acid/H+ QWC: Equilibrium shifts forming HCOOH OR (HCOOH) Equilibrium left
A chemist prepares a buffer solution by mixing together the following:200 cm3 of 3.20 mol dm–3 HCOOH (Ka = 1.70 × 10–4 mol dm–3) and800 cm3 of 0.500 mol dm–3 NaOH.The volume of the buffer solution is 1.00 dm3. • Explain why a buffer solution is formed when these two solutions are mixed together. • Calculate the pH of this buffer solution. Give your answer to two decimal places.
HCOOH reacts with NaOH forming HCOO– /HCOONa OR HCOOH + NaOH HCOONa + H2O Equilibrium sign allowed (Some) HCOOH/(weak) acid remains OR HCOOH/(weak) acid is in excess n(HCOOH) OR [HCOOH] = 0.24(0) (mol / mol dm–3) n(HCOO–) OR [HCOO– ] OR [HCOONa] = 0.4(00) (mol / mol dm–3) [H+] = K a [HCOOH] [HCOO– ] pH = –log [H+] = –log(1.70 10–4 0.24 0.4 ) = 3.99
Write the ionic equation for the reaction between aqueous butanoic acid and magnesium.
Mg + 2H+ Mg2+ + H2