Question

1) 20. mL sample of 0.20 M HOCl is added to 25.0mL of 0.20M KOH. What is the pH of the solution? -8​ HOCl (K​a​ = 3.5 x 10​ )

2) The ideal pH for the pool is 7.4. What is the ratio of (OCl-​ ​) to (HOCl) needed to maintain the swimming pool at a pH of 7.4?

3) What would be the new pH if you added 0.3 moles of NaOH to 1.0L of the pool water at pH = 7.4? Assume you have 1.0 mole of HOCl

with steps please I need to know how to do these for my exam tomorrow

Answer 1

1)

Given:
M(HOCl) = 0.2 M
V(HOCl) = 20 mL
M(KOH) = 0.2 M
V(KOH) = 25 mL


mol(HOCl) = M(HOCl) * V(HOCl)
mol(HOCl) = 0.2 M * 20 mL = 4 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.2 M * 25 mL = 5 mmol


We have:
mol(HOCl) = 4 mmol
mol(KOH) = 5 mmol

4 mmol of both will react

excess KOH remaining = 1 mmol
Volume of Solution = 20 + 25 = 45 mL
[OH-] = 1 mmol/45 mL = 0.0222 M

use:
pOH = -log [OH-]
= -log (2.222*10^-2)
= 1.6532


use:
PH = 14 - pOH
= 14 - 1.6532
= 12.3468
Answer: 12.35

Only 1 question at a time please