Question

Mix 0.05M KH2PO4 and 0.05M K2HPO4 in the following volumes (ml) in 100mL beakers or cups:

**this only like 5 cups to save time**

Cup	K2HPO4 (ml)	KH2PO4(ml)
1	0	50
2	1	49
3	5	45
4	10	40
5	20	30

For each cup calculate these four values:

Concentration of K2HPO4, Concentration of KH2PO4, % of total phosphate in the form of K2HPO4 and log[K2HPO4/KH2PO4]

Please show your work for each calculation!! you can just solve for one cup thats fine i just dont understand how to solve for these.

Answer 1

Concentration of solutions before mixing.

Cup 1

0 mL of KH2PO4 + 50 mL of K2HPO4.

Hence, since there is no K2HPO4 added,

Only added solution is 50 mL of K2HPO4. Hence, there will be no concentration change. Hence,  .

percentage of total phosphate in form of K2HPO4 = 0% since there is no K2HPO4 in the solution.

Cup 2

1 mL K2HPO4 + 49 mL KH2PO4

Number of moles of K2HPO4 in 1 ml of 0.05 mol/L solution

Final volume = 1mL + 49 mL = 50 mL

Hence, concentration of K2HPO4 is

Similarly, number of moles of KH2PO4 in 49 mL of 0.05 M solution is

Concentration of KH2PO4 is

percentage of total phosphate in K2HPO4 =

Cup 3

5mL K2HPO4 + 45 mL KH2PO4

Number of moles of K2HPO4 in 5 ml of 0.05 mol/L solution

Final volume = 50 mL

Hence, concentration of K2HPO4 is

Similarly, number of moles of KH2PO4 in 45 mL of 0.05 M solution is

Concentration of KH2PO4 is

percentage of total phosphate in K2HPO4 =

Cup 4

10 mL K2HPO4 + 40 mL KH2PO4

Number of moles of K2HPO4 in 10 ml of 0.05 mol/L solution

Final volume = 50 mL

Hence, concentration of K2HPO4 is

Similarly, number of moles of KH2PO4 in 40 mL of 0.05 M solution is

Concentration of KH2PO4 is

percentage of total phosphate in K2HPO4 =

Cup 5

20 mL K2HPO4 + 30 mL KH2PO4

Number of moles of K2HPO4 in 20 ml of 0.05 mol/L solution

Final volume = 50 mL

Hence, concentration of K2HPO4 is

Similarly, number of moles of KH2PO4 in 30 mL of 0.05 M solution is

Concentration of KH2PO4 is

percentage of total phosphate in K2HPO4 =