Saturday 6th September
TODAY WE ARE
LEARNING ABOUT
 What can we learn from titrations?
TODAY'S
KEY WORDS ARE
  • _W_D_scr_b_  h_w  t_  c_rry  __t  t_tr_t__ns  _s_ng  str_ng  _c_ds  _nd  str_ng  _lk_l_s  _nly  (s_lf_r_c,  hydr_chl_r_c  _nd  n_tr_c  _c_ds  _nly)  t_  f_nd  th_  r__ct_ng  v_l_m_s  _cc_r_t_ly_
  • C_nc_ntr_t__n
  • m_l_
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    Memory Anchor:

    YOU WILL SHOW
    YOUR LEARNING BY...

    • Super Challenge:

      Stretch:

      Challenge:

    Key Questions:

    1. A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? (The relative formula mass, Mr, of ethanoic acid is 60)
      • concentration in g/dm3 = concentration in g/dm3 ? Mr concentration = 0.1 ? 60 = 6 g/dm3 Answer 6 g/dm3
    2. 25 cm3 of dilute hydrochloric acid is neutralised by 20 cm3 of 0.5 mol/dm3 sodium hydroxide. What is the concentration of the hydrochloric acid?
      • Step 1: Convert volumes to dm3 25 cm3 of HCl = 25 /1000 = 0.025 dm3 20 cm3 of NaOH = 20 / 1000 = 0.020 dm3 Step 2: Determine the number of moles of sodium hydroxide moles of NaOH = concentration ? volume moles of NaOH = 0.5 x 0.020 = 0.010 mol Step 3: Work out the number of moles of acid using the balanced equation HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l) In this reaction, one mole of HCl reacts with one mole of NaOH. This is a 1:1 ratio. Therefore, in our titration, 0.010 mol of NaOH must neutralise 0.010 mol of HCl. Step 4: Calculate the concentration of the acid concentration of HCl = number of moles / volume concentration of HCl = 0.010 / 0.025 = 0.4 mol/dm3 Answer The concentration of the HCl is 0.4 mol/dm3.