Question

Incorrect Question 5 0/1 pts A 25.0 mL sample of 0.723 M HCIO4 is titrated with 0.27 M KOH solution. The H30* concentration after the addition of 80.0 mL of KOH is M. (Cl is chlorine, not carbon and iodine) 0.4 3x10-13 1x107 0.7 4x10-2
Please explain the correct answer for both questions. Thank you!

Answer 1

Answer 5

Correct answer is b)

Explanation:

Moles of HClO_{​​​​​4}= 0.723 M*25 mL = 18.075 mmol

Moles of KOH = 0.270 M*80 mL = 21.600 mol

As moles of KOH are more, so, excess moles of KOH = (21.600 - 18.075) millimol

= 3.525 millimol

Total volume = 25 ml + 80 ml = 105 ml

[OH^-] = 3.525 millimol/105 ml = 0.0336 M

[H⁺] = 10^-14/0.0336 = 29.76 x 10^-14 M

= 3.0 x 10^-13 M

Answer 6

Correct answer is d) 7

Explanation: pKa is the pH at the middle region of least steep curve.