[Exitus]

1.
a)You put [H2] = 0,70M, [I2] = 0,70M and [HI] =1,50M. And the K value is 49. Decide the equilibrium concentrations. H2 + I2 <-> 2HI

b) Which concentration HI gets inserted if the iodine concentration at a new equilibrium is 0,42Molar. The K value is still 49.

c) What happens with the equilibrium if the volume increases (Nothing? Since it's 2 moles on both sides of the reaktion)
 
2. What's the concentration of HAc(acetic acid) with pH = 3. the Ka value is 1,7*10^-5 -

3. Calculate pH in a mixture of 0,100M HCL(hydrochloric acid) and 0,250M Ca(OH)2 (calcium hydroxide) with the total volume of 500,0 cm^3 (cubic centimeters)

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

[Exitus]

Alright! I'll post my answers soon.

[Exitus]

You have to show your attempts at solving the question to receive help. This is a forum policy.

Question 3:
Answer: pH =13,6
Solution: n(HCL) = 0,1*0,5 = 0,05moles -> n(H+) = 0,05moles
n(Ca(OH)2)= 0,250*0,5 = 0,125moles -> n(OH-) = 0,125*2 = 0,250moles
neutralisation gives us 0,250-0,05 = 0,200 moles n(OH-)
and [OH-] = 0,200/0,5 = 0,4
-log0,4 = 0,4 = pOH
pH = 14-0,4 = 13,6

[Borek]

To be honest I have no idea how to understand the 3rd problem. Mixture with the final volume can mean anything starting with just 500 mL of the initial HCl solution, ending with the 500 mL of the initial Ca(OH)₂ solution.

[Exitus]

To be honest I have no idea how to understand the 3rd problem. Mixture with the final volume can mean anything starting with just 500 mL of the initial HCl solution, ending with the 500 mL of the initial Ca(OH)₂ solution.

Ok but

Question 2:
Answer: 0,059M
Solution: pH 3 gives us [H+] = 10^-3
 HAc <-> H+ + Ac-
Concentration before equilibrium: [HAc] : 0 : 0
Change: -10^-3 : +10^-3 : +10^-3
Concentration after equilibrium [HAc] - 10^-3 : 10^-3 : 10^-3
Here I neglect the -10^-3 from [HAc] since it's less than 10%
1,7*10^-5 = 10^-6/[HAc] which gives [HAc] = 0,059M