9 solution calculations

• Steps to complete solutions calculations

  1. Use the volume and concentration of one reactant to calculate the moles.
  2. Use the balanced (or ionic) equation to find the moles of the other reactant
  3. Calculate the volume or concentration as required of that reactant
• Concentration = moles ÷ volume (dm-³)
• Volume in dm³ = volume in cm³ ÷ 1000 e.g 25cm³ -> 25 ÷ 1000 = 0.025dm³)
• Concentration in gdm-³ = concentration in mol dm-³ × Mr (e.g H2SO4 0.10 moldm-³ -> 0.10 × 98dm³ = 9.8gdm-³)
• In many titrations, a standard solution of one the reagents is made (typically 250cm³ in volumetric flask) and 25cm³ portions of this standard solution are used in each titration
• Monoprotic acid means one H+ ion per unit and examples are HCl, HNO3, CH3COOH. The reacting ratio with NaOH is 1:1 (acid : NaOH)
• Diprotic acid means two H+ ions per unit and examples of this are H2SO4 and the reacting ratio with NaOH is 1:2 (acid : NaOH)
• Triprotic acid means three H+ ions per unit and examples are H3PO4 and the reacting ratio with NaOH is 1:3 (acid : NaOH)
• Example of solution calculation

  25.0cm³ of 0.020moldm-³ sulfuric acid neutralises 18.6cm³ of sodium hydroxide solution

  H2SO4(aq) + 2NaOH(aq) -> Na2SO4(s) + 2H20(l)

  A)find the concentration of the sodium hydroxide solution in moldm-³

  Moles of H2SO4 = conc x vol =

  0.020 × (25÷1000) = 0.000500

  Moles of NaOH = conc x vol = 2 x Moles H2SO4 = 0.000500 x 2 = 0.00100

  Concentration of NaOH = Moles ÷ volume = 0.00100 ÷ (18.6÷1000) = 0.0538moldm-³

  B) find the concentration of th sodium hydroxide solution in gdm-³

  Mr of NaOH = 23 + 16 + 1 = 40

  Mass of NaOH in 1dm-³ = Mr x Moles = 40 x 0.0538 = 2.15g

  Concentration = 2.15gdm‐³
• Another example of solution calculation

  Crystals of citric acid contain water of crystallisation (C6H8O7.nH2O), Citric and is a triprotic acid 1.52 g of the citric acid was made up to 250 cm³ solution. 25 cm³ portions of this solution required 21.80 cm³ of 0.100 mol dm-³ sodium hydroxide for neutralisation Calculate the Mr. and the value of n 0.00218 Moles of NaOH = conc x vol (dm³) = 0.100 x (21.80 ÷ 1000) = 0.00218

  Moles of C6H8O7.nH2O in each titration = 0.00218 ÷ 3 = 0.000727 (1mol of acid reacts with 3mol of NaOH)

  Moles of C6H8O7.nH2O in 250 cm³ solution = 0.000727 x 10 = 0.00727

  Mr of C6H8O7.nH2O = mass = 1.52÷0.00727 = 209.2

  Mr of nH2O = 209.2-192.0=17.0

  n = 17 ÷ 18 = 0.954 = 1 (n is a whole number)
• A back titration is done to analyse a base (or acid) that does not react easily or quickly with an acid (or base). As an alternative the base (or acid) is treated with an excess of acid (or base) and then the left over acid (or base) titrated. You can then work back to find out about the original base (or acid)
• Examole of a back titration

  Imagine that we are trying to find out how many moles of CaCO3 we have (let's call it p moles). We add 10 moles of HCI (an excess). The excess is made into a 250 cm³ solution in a volumetric flask and then 25cm³ portions of it require 0.4 moles of NaOH for neutralisation CaCO3 + 2HCl -> CaCl2 + H₂O + CO2

  HCI NaOH NaCl HO

  This means that there is 10 × 0.4 moles (= 4 moles) of left over HCI in the volumetric flask solution (as we used 1/10 of the solution in the volumetric flask in each titration) This means that 6 moles (10-4 moles) of HCI reacted with the CaCO3 This means that there must have been 3 moles of CaCO3 (i.e. p = 3 ) in the first place (remember that 2 moles of HCI reacts with each mole of CaCO3).