For the titration of 10.0 mL of o.200 M nitrous acid HNO2 with 0.100 M sodium hy
ID: 692028 • Letter: F
Question
For the titration of 10.0 mL of o.200 M nitrous acid HNO2 with 0.100 M sodium hydroxi (Please use pK, 3.34 for nitrous acid). 4. ide, NaOH, answer the following questions and construct a titration curve. a. What volume (in mL) of 0.100M sodium hydroxide is required to "neutralize" 10.0 mL of 0.200M nitrous acid? b. What is the pH of the solution before any sodium hydroxide is added? c. What is the pH at the half-stoichiometric or half-equivalence point (that is, what is the pH when you have added of the total volume of base that would be required to react with all of the acid originally present)? What is the pH when 95% of the sodium hydroxide required to reach the stoichiometric point has been added? d. e. What is the pH at the stoichiometric point? What is the pH when a 5% excess of the sodium hydroxide required to reach the stoichiometric point has been added?Explanation / Answer
a)
volume of NaOH for neutralization
ratio is 1 mol of acid = 1 mol of base
mmol of acid = MV = 10*0.2 = 2
mmol of base = 2
Vbase = mmol base/Mbase =2/0.1 = 20 mL of base required
b)
pH before addition of base
First, assume the acid:
to be HA, for simplicity, so it will ionize as follows:
HA <-> H+ + A-
where, H+ is the proton and A- the conjugate base, HA is molecular acid
Ka = [H+][A-]/[HA]; by definition
initially
[H+] = 0
[A-] = 0
[HA] = M;
the change
initially
[H+] = + x
[A-] = + x
[HA] = - x
in equilbrirum
[H+] = 0 + x
[A-] = 0 + x
[HA] = M - x
substitute in Ka
Ka = [H+][A-]/[HA]
Ka = x*x/(M-x)
x^2 + Kax - M*Ka = 0
if M = 0.2 M; then
x^2 + (10^-3.34)x - 0.3*(10^-3.34) = 0
solve for x
x =0.00933
substitute
[H+] = 0 + 0.00933= 0.00933M
pH = -log(H+) = -log(0.00933) = 2.03
c)
in half equivalence point
mmol of HNO2 left = 1
mmol of NO2- formed = 1
this is a buffer so
pH = pKa + log(NO2-/HNO2)
pH = 3.34 + log(1/1)
pH = 3.34
d)
at 95% point
mmol of HNO2 left = 2*0.05 = 0.10
mmol of NO2- formed = 2*0.95 = 1.90
pH = pKa + log(NO2-/HNO2)
pH = 3.34 + log(1.90/0.1) = 4.6187
e)
pH in equiv. point
expect hydrolysis
since A- is formed
the next equilibrium is formed, the conjugate acid and water
A- + H2O <-> HA + OH-
The equilibrium s best described by Kb, the base constant
Kb by definition since it is an base:
Kb = [HA ][OH-]/[A-]
get ICE table:
Initially
[OH-] = 0
[HA] = 0
[A-] = M
the Change
[OH-] = + x
[HA] = + x
[A-] = - x
in Equilibrium
[OH-] = 0 + x
[HA] = 0 + x
[A-] = M - x
substitute in Kb expression
Kb = [HA ][OH-]/[A-]
Ka can be calculated as follows:
Kb = Kw/Ka = (10^-14)/(10^-3.34) = 2.18*10^-11
2.18*10^-11= x*x/(0.1-x)
solve for x
x^2 + Kb*x - M*Kb = 0
solve for x with quadratic equation
x = OH- =1.47*10^-6
[OH-] =1.47*10^-6
pOH = -log(OH-) = -log(1.47*10^-6 = 5.83
pH = 14-5.83= 8.17
pH = 8.17
f)
excess base
vbase = 1.05*20 = 21 mL
Vtotal = 20+21 = 41 mL
mmol of OH- left = 1*0.1 = 0.1
[OH-] = mmol/V = 0.1/41 = 0.00243
pH = 14 + log(OH) = 14 +log(0.00243) = 11.39
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