sorry posting almost when day is over :P
1)Calculate the concentrations of the non-ionised acid of the ions in a 0.1 M formic acid at equilibrium. ( Ka =1.7* 10^-4)
2)A certain weak acid has Ka=10^-4 Calculate the equilibrium constant for its reaction with a strong base.
3)100mL of 0.1N NaOH is mixed with 100ml of 0.1N CH3COOH . What is the pH of the resulting solution. Given Ka = 1.8*10-5
4)The solubility of calcium hydroxide in water is 1.63 g per litre . 500mL of a saturated solution of calcium hydroxide is shaken with 500mL of 0.2M sodium hydroxide. Determine the mass of calcium hydroxide that separates out as a precipitate.
5) The standard potential of the Ag+ Ag is 0.8V. A silver wire is kept immersed in a saturated solution of AgI whose solubility product constant is 910-17 . Calculate the potential developed at the electrode.
6)What is the pH of the solution when 0.01 moles of HCl is added to 500mL of a solution containing 0.25M acetic acid and 0.25M sod acetate? Assume no change in volume due to HCl addition.
{Ka(CH3COOH) = 1.8 x 10 -5.}
106Q1. Calculate the concentrations of the non-ionised acid of the ions in a 0.1 M formic acid at equilibrium. ( Ka =1.7* 10^-4)
HCOOH = HCOO- + H+
0.1 0 0
0.1(1-a) 0.1a 0.1a
Ka = 0.01a20.1(1-a) = 0.1a21-a
We can solve for a and then calculate 0.01(1-a)
Here ostwald formula cant be used bcz a is NOT << 1
106Q3.100mL of 0.1N NaOH is mixed with 100ml of 0.1N CH3COOH . What is the pH of the resulting solution. Given Ka = 1.8*10-5
NaOH + CH3COOH = CH3COONa + H2O
10 10
millimole millimole
So, the product will consist of CH3COONa (10 millimole) in 200 ml, so conc = 0.05
So, CH3COO- + H2O = CH3COOH + OH-
0.05
0.05(1-a) 0.05a 0.05a
Keq = Kb = 25*10-4a20.05(1-a) = 5*10-2a21-a
Solve for a and find [OH-] then pH can be found
243)
NaOH +CH3COOH →CH3COONa +H2O
10mol 10mol 0 0
0 0 10 10
conc acetate=10/200=1/20=0.05
pH=1/2 [14-4.7+log 0.05]=1/2[9.3+0.7-2]=1/2(8)=4
242)
HA →H++A-
K=[H+] [A-] /[HA] =10-4
Now reaction
HA+OH-→H2O +A-
K'=[A-]/[HA][OH-]
=10-4/10-14=1010
