Question

I would like help understanding these questions.

Acids/Bases 1. Calculate the volume (in mL) of a 1.75 M KOH solution required to completely titrate (neutralize) 16.0 mL of a 3.00 M phosphoric acid solution. 2. Calculate the concentration of a sodium hydroxide solution that required 43.40 mL to completely titrate 26.50 mL of a 0.350 M oxalic acid (H2C2O4) solution. 3. Calculate the pH of the following: (a) a solution containing a hydronium concentration of 0.150 M (b) a 0.75 M HCl solution 4. Calculate the concentration of H* ions in solution if the pH is determined to be 4.70.

Thank you for your time!

Answer 1

1) Consider a reaction, H₃PO₄ + 3 KOH K₃PO₄ + 3 H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base = 1/3

We have correlation, M _acid V _acid = M _base V _base stoichiometric ratio.

Therefore, 3.0 M   16.0 ml =1.75 M V _base (1/3)

V _base = 3.0 M   16.0 ml / 1.75 M (1/3)

V _base = 82.28 ml

ANSWER: Volume of KOH solution required to completely neutralize 16.0 ml of 3.0 M phosphoric acid solution is 82.28 ml.

2)

Consider a reaction, H₂C₂O₄ + 2 NaOH Na₂C₂O₄ + 2 H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base = 1/2

We have correlation, M _acid V _acid = M _base V _base stoichiometric ratio.

Therefore, 0.350 M   26.50 ml =M _base 43.40 ml   (1/2)

M _base = 0.350 M   26.50 ml / 43.40 ml   (1/2)

M _base = 0.427 M

ANSWER: [ NaOH] = 0.427 M

3)

a) Given : [H₃O ⁺ ] =0.150 M

We have, pH = - log [H₃O ⁺ ] = - log 0.150 = 0.824

b) HCl is a strong acid. It dissociates completely into ions when dissolved in water.

Consider dissociation of HCl in water.

HCl (aq) + H₂O (l) $\rightarrow$ H₃O ⁺ + Cl ^-

From reaction, 1 mol HCl $\equiv$ 1 mol H₃O ⁺

$\therefore$ [HCl] = [H₃O ⁺ ] = 0.75 M

We have, pH = - log [H₃O ⁺ ] = - log 0.75 = 0.125

C) We have, pH = - log [H₃O ⁺ ]

$\therefore$ [H₃O ⁺ ] = 10 ^{- pH} = 10 ^{- 4.70} = 2.00