Question

Calculate pH of 0.1 M potassium hydrogen oxalate. pK_a1=1.25, pK_a2=4.27.

Calculate pH of solution prepared from 0.1 mole of formic acid and 0.02 mole of NaOH diluted to 1L. pK_a=3.75.

You have 100 mL of 1 M ammonia solution (pK_a=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.5?

What will be pH of a 100 mL buffer solution containing 0.064 M of acetic acid and 0.036 M of sodium acetate, after addition of 1 millimole of the hydrochloric acid? pK_a=4.75.

Explain what happen to man who have an acidic blood fluid in his body.

Answer 1

Question 1:

Potassium hydrogen oxalate is an amphoteric salt and the pH of an amphoteric salt is given by the equation

pH = 1/2 x (pK_a1 + pK_a2) ------------------ (1)

here,

pK_a1 = 1.25

pK_a2 = 4.27

So according to equation (1) we get

pH = (1/2) x (1.25 + 4.27) = (1/2) x 5.52 = 2.76

Answer: pH of 0.1M potassium hydrogen oxalate = 2.76

Question 2:

Formic acid and NaOH will react to form sodium formate as per the below given equation

HCOOH + NaOH HCOONa + H₂O

We have taken 0.1 mol of formic acid and 0.02 moles of NaOH.

Here NaOH will act as a limiting reagent ( i.e. it will be consumed completely in the reaction ) and the amount of the product, HCOONa, formed will be equal to that of the concentration of the limiting reagent. The solution left behind will be a buffer solution as it contains only a weak acid (HCOOH) and its salt (HCOONa)

So,

concentration of HCOONa (salt) formed = 0.02moles -----------------(2)

( 0.02moles of HCOOH + 0.02moles of NaOH 0.02moles of HCOONa )

We were having 0.1mol of HCOOH and out of it 0.02 moles will react to form HCOONa

i.e. the remaining acid present in the solution = 0.1 - 0.02 = 0.08moles ---------(3)

i.e. as per equation (2) and (3)

[ Salt ] = 0.02 moles/L (Since solution diluted to 1L) = 0.02M

[ Acid ] = 0.08moles/L (Since solution diluted to 1L) = 0.08M

Now by applying the Henderson- Hasselbalch equation

pH = pKa + log ([Salt] / [Acid]) ----------------- (4)

i.e. pH = 3.75 + log (0.02/0.08) = 3.14

Answer: pH of the solution = 3.14

Question 3:

pH + pOH = 14

So, pOH = 14-9.5 = 4.5

Similarly

pKa + pKb = 14

So, pKb = 14 - 9.25 = 4.75

Now by applying the Henderson- Hasselbalch equation for weak base and its salt

pOH = pK_b + log ([Salt] / [Base])

i.e. 4.5 = 4.75 + log ([Salt] / [Base])

log ([Salt] / [Base]) = -0.25

([Salt] / [Base]) = 0.5623

i.e [n_HCl / V_HCl] /([n_NH3/V_NH3] - [n_HCl/V_HCl]) = 0.5623

[n_HCl / V_HCl] /(0.01 - [n_HCl/V_HCl]) = 0.5623

i.e. 1.5623 C_HCl = 0.05623

C_HCl = n_HCl / V_HCl = 0.05623

C_HCl = n_HCl / V_HCl = 0.05623

1.5623 C_HCl = 0.05623

CHCl = 0.0359

i.e. VHCl = nHCl/CHCl2.79

Question 4:

Given:

concentration of Acetic acid = 100ml x 0.064M = 6.4mmole

Concentration of Sodium acetate = 100ml x 0.036M = 3.6mmole

Now to this solution we are adding 1mmole of HCl. HCl being a strong acid will dissociate completely to give H⁺ ions and these H⁺ ions will be released to the solution.

i.e. amount of H⁺ ions added = 1mmole

These H⁺ ions will result in the reaction CH₃COO^- + H⁺ CH₃COOH

i.e amount of acetic acid will increase and that of the salt, sodium acetate will decrease correspondingly by 1mmole

Therefore after adding HCl,

[ Acetic acid ] = 6.4 + 1 mmole = 7.4mmole in 100ml of solution

[ Sodium acetate ] = 3.6 - 1 mmole = 2.6mmole in 100ml of solution

100 ml contain 7.4mmol acetic acid 1000ml contain 74 mmol/L = 0.074M -----------------(5)

similarly 100 ml contain 2.6mmol sodium acetate 1000ml contain 26 mmol/L = 0.026M--------------(6)

Now by applying the Henderson- Hasselbalch equation and considering equation (5) and (6), we get

pH = pKa + log ([Salt] / [Acid]) = 4.75 + log (0.026/0.074) = 4.29

Answer: pH of the solution = 4.29