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Acids ph buffers A2 chemistry. Using the post it notes write down the name of the species and whether each species round the room is an acid, base or.

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Presentation on theme: "Acids ph buffers A2 chemistry. Using the post it notes write down the name of the species and whether each species round the room is an acid, base or."— Presentation transcript:

1 Acids ph buffers A2 chemistry

2 Using the post it notes write down the name of the species and whether each species round the room is an acid, base or alkali

3 H3O+H3O+ pka = -1.7

4 H2OH2O pka = 15.7

5 HI pka = -10

6 NH 4 + pka = 9.2

7 Al 2 O 3

8 Na 2 O

9 SO 2

10

11 Ammonia pka = 36

12 Complete the equations match up pka = 36

13 Loop game

14 0.8L of 1M ethanoic acid reacted was titrated with 0.2L 1M NaOH. Calculate the pH of the solution. Ka = 1.76 x Acid moles [HA]H+H+ Base moles Initial change Equilibrium Step 1. Work out moles of HA at start Step 2. Work out moles of base at start Step 3. Work out moles of HA that reacted with base – this is moles H+ Step 4. Minus moles of neutralised HA from initial acid to get the moles of HA Ka x [HA] = [H+] 2 Step 5. Use equilibrium concentrations and put them in this equation  Step 6. use pH = -log [H+]

15 0.25L of 1M ethanoic acid reacted was titrated with 0.25L 1M Ca(OH) 2. Calculate the pH of the solution. Ka = 1.76 x Acid moles [HA]H+H+ Base moles Initial change Equilibrium Step 1. Work out moles of HA at start Step 2. Work out moles of base at start Step 3. Work out moles of HA that reacted with base – this is moles H+ Step 4. Minus moles of neutralised HA from initial acid to get the moles of HA Ka x [HA] = [H+] 2 Step 5. Use equilibrium concentrations and put them in this equation  Step 6. use pH = -log [H+]

16 0.3L of 1M Methanoic acid reacted was titrated with 0.7L 1M Ca(OH) 2. Calculate the pH of the solution. pKa = 3.77 Acid moles [HA]H+H+ Base moles Initial change Equilibrium Step 1. Work out moles of HA at start Step 2. Work out moles of base at start Step 3. Work out moles of HA that reacted with base – this is moles H+ Step 4. Minus moles of neutralised HA from initial acid to get the moles of HA Ka x [HA] = [H+] 2 Step 5. Use equilibrium concentrations and put them in this equation  Step 6. use pH = -log [H+]

17 Q1

18 A1

19 Q2

20 Q3 Q4

21 A2 A3

22 A4

23 Q5

24 Q6

25 Q7

26 Q8

27 A5

28 A6

29 A7

30 A8

31

32 Q9

33 Q10

34 A9 A10


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