2. Rough1234 Initial0.015.730.60.215.3 Final15.730.646.215.330.6 Total Should another titration be done? Why or why not?

3.  Find the average titre volume.  15.1 +15.1+15.2 /3=15.1mL  3 results within.1mL are called concordant results  (exclude trial 1 since it is more than.1 away from the rest) Rough1234 Initial0.015.730.60.215.3 Final15.730.645.715.330.5 Total15.714.915.1 15.2

4.  Remember that c=n/V  c=concentrantion  n=moles  V=volume (in L)

5. 1. Identify the known and the unknown in the problem. 2. Find the amount of moles of the known using the c=n/V equation. 3. Look at the balanced equation to find the mole ratio and multiply the moles of the known by the mole ratio (unknown/known) to find the moles of the unknown. 4. The concentration of the unknown can be calculated from the moles calculated above and the volume from the titration.

6. 15.0 mL of an aqueous solution of 0.105 mol L -1 sodium carbonate is titrated with hydrochloric acid solution.12.6mL of solution is needed to reach the equivalence point. What is the concentration of the sodium carbonate solution? 2 HCl (aq) + Na 2 CO 3(aq)  2 NaCl (aq) + CO 2(g) + H 2 O (l)

7. HClNa 2 CO 3 ratio21 n c0.105 mol L -1 V0.01260.015L

8.  Calculate moles of known – sodium carbonate  n=c*V  n=0.105*0.015L  n=0.001575mol

9. HClNa 2 CO 3 ratio21 n0.00158 mol c0.105 mol L -1 V0.0126L0.0150L

10. Find the number of moles of unknown using mole ratio and moles of known. =0.001575mol *(unknown/known) =0.001575mol * 2/1 =0.00315mol of HCl

11. HClNa 2 CO 3 ratio21 n0.00315m ol 0.00158 mol c0.105 mol L -1 V0.0126L0.0150L

12.  Calculate concentration of unknown  c=n/V  c=0.00315mol/0.0126L  c=0.250mol L -1

13. HClNa 2 CO 3 ratio21 n0.00315mo l 0.00158 mol c0.250 mol L -1 0.105 mol L - 1 V0.0126L0.0150L

14.  2HCl (aq) + Ca(OH) 2(aq)  CaCl 2(aq) + 2H 2 O (l)  We know the following…  V = 50.0 mL Ca(OH) 2 in flask  [HCl] = 0.100 mol L -1  V HCl added = 20.0 mL for 3 concordant titres.