Question

Consider a 0.10 M solution of HNO₂ that was Ka = 4.5x10^-4. A researched was tasked to make a 100 mL solution of buffer by Titration with 0.10 M NaOH

A)What is the initial ph of the solution

B)The researched misread the instruction and erroneously added 50.0 mL of NaOH solution to 50.0 mL of the HNO₂ solution. What is the pH of that solution

C)The researched reared the instruction but still did not know what to do and therefore decided instead to simply add 0.500 g of NaNO₂ to a 100 mL of HNO₂ solution and wish for luck. What is the pH of this solution? Note that the molar mass of NaNO₂ is 69.0 g/mol

D)Help this researched and determine what actual volumes of HNO₂ and NaOH were needed to make 100. mL of this buffer.

Answer 1

A)

0.10 M solution of HNO₂

Ka = 4.5x10^-4.

HNO₂ +H₂O <----> H₃O⁺ + NO₂^-

I 0.10                      0         0

C –x                        +x       +x

E 0.10 – x               x         x

Ka =[H₃O^+1] [NO₂^ -1]/ [HNO₂]

4.5 X 10^-4 = [X] [X] / [0.10-x]

Since Ka is very small, we will assume the amount of HNO₂ dissociated is very small, so ignore the -x. .

4.5 X 10^-4 = [X] [X] / [0.10]

(4.5 X 10^-4) ([0.10]) = X^2

X = 6.70 x 10^-3 =[H3O^+1]

pH = - log H⁺

pH = - log 6.70 x 10^-3

pH = 2.17

B)

50.0 mL of 0.10 M NaOH solution

So moles of NaOH = 0.050 L x 0.10M = 0.005 moles

50.0 mL of 0.10 M the HNO₂ solution

So, moles of HNO₂ = 0.050 L x 0.10M = 0.005 moles

At this point, the pH = pKa, because the concentrations of base and acid are equimolar. It is 1/2 equivalence point.

The Henderson-Haselbalch equation is used for pH calculation of a solution containing pair of acid and conjugate base.

PH = PKa + log [base]/[acid]

At half equivalence point, the concentrations of the weak acid and its conjugate base are equal.

So the ([A-]/[HA]) ratio is 1.

The log of 1 is zero, so, the pH = pKa

pKa = - log Ka = - log (4.5 X 10^-4 )

pKa = 3.35 = pH

C-

0. 500 g of NaNO₂

The molar mass of NaNO₂ is 69.0 g/mol

Moles of NaNO₂ = 0. 500 g/ 69.0 g/mol = 0.00725 moles

Molarity of NaNO₂ = 0.00725/0.1 = 0.0725M

Moles of HNO2 = 0.050 L x 0.10M = 0.005 moles

Molarity of HNO2 = 0.005/0.1 = 0.05 M

PH = PKa + log [base]/[acid]

PH = 3.35 + log (0.0725/0.05)

PH = 3.35 + 0.16 = 3.51

D)

PH = PKa + log [base]/[acid]

3.51 = 3.35 + log [base]/[acid]

log [base]/[acid] = 3.51 - 3.35

log [base]/[acid] = 0.16

[base]/[acid] = 1.17/1

Ratio between volumes of HNO₂ and NaOH = 1.17:1

To make 100 mL of this buffer, volume of HNO₂ = (1/2.17) x 100 ml = 46 ml

Actual volume of NaOH = 100ml – 46ml = 54 ml