1) A solution of Na2CO3 has a pH of 10. The CO32- on is the conjugate base of th
ID: 826374 • Letter: 1
Question
1) A solution of Na2CO3 has a pH of 10. The CO32- on is the conjugate base of the HCO3- ion. Write the net ionic equation for the reaction which makes a solution of Na2CO3 basic. ( My book refers me to this equation: B^- (aq) + H^+ (aq) -----> HB (aq) )
2) Formic acid, HFor, has a Ka equal to about 1.8 x 10^-4. A student is asked to prepare a buffer having a ph of 3.4 from a solution of formic acid and a solution of sodium formate having the same molarity. How many milliliters of the NaFor solution should she add to 20 mL of the HFor solution to make the buffer?
3) How many mL of 0.10M NaOH should the student added to 20 mL 0.10M HFor if she wished to prepare a buffer with a pH of 3.4, the same as in Problem 1?
Explanation / Answer
1)
CO3^2- + H2O ---> HCO3- + OH-
increase OH- increases the pH
2)
From Henderson Hasselblach equation
pH = pKa + log [A-]/[HA]
where
[A-]= molar concentration of a conjugate base
[HA] = molar concentration of a undissociated weak acid
pKa = -log ka = -log 1.8 x 10-4 = 3.75
pH = 3.4
and [A-] =[HA]
Hence [AH] = 20 x where x is molairity and [A-] = V x where V is the volume of NaF
Therefore,
3.4 = 3.75 + log Vx/20 x
=> -0.35 = log V -log 20
=> log V = 1.30 -0.35
=> log V = 0.95
=> V = 8.91
Required volume of NaF to be added 8.91 mL
3)
NaOH --> Na+ + OH-
HCOOH <--> HCOO- + H+
in this case when we add x molar of NaOH it dissociate to Na+ and OH-, then OH- will consume H+ from HCOOH and will produce water. by the Le-chatelier's principle the HCOOH reaction goes to right side to produce more H+; from here HCOO- will produce.
it means 1 mol NaOH will produce 1mol Na+ & 1mol OH-
1mol OH- consumes 1mol H+ and HCOOH and produces 1mol HCOO-
by this explanation; finally the concentration of Na+ and HCOO- in the solution should be the same;
because we want the pH of buffer 3.4 as before then such as previous problem
??[HCOO-] / [HCOOH ] = 0.457
no of mmoles of Na+ = 0.10 * V
no of mmoles of HCOOH = 20 * 0.10 - 0.1*V
{ 0.10 * V / (20+V) } / {(20 * 0.10 - 0.1*V) / (20+V)} = 0.457
V =6.27 mL
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