Question

19.What quantity of NaOH(s) must be added to 1.00 L of 0.200 M HCl to achieve a pH of 12.00? (Assume no volume change.)

20.A 65.5-mL sample of 0.14 M HNO2 (Ka = 4.0 x 10–4) is titrated with 0.11 M NaOH. What is the pH after 26.8 mL of NaOH has been added?

21.

What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00?

Ka for HOCl is 3.5 x 10–8.

Answer 1

1.

moles of HCl = (0.200 mol/L) (1.00 L) = 0.200 mol

So, 0.200 moles of NaOH gets us to pH = 7

pH = 12 means pOH = 2 means [OH-] = 0.01 M

1.00 L of pH = 7 solution requires 0.01 mol of NaOH to get to pH = 12

part a answer is 0.201 moles of NaOH.

I assume part b means the NH3 solution

M1V1 = M2V2

(0.500) (100 mL) = (0.500) (x)

x = 100 mL

2.

Moles HNO2 = 0.0500 L x 0.10 M = 0.00500
Moles OH- = 0.0250 L x 0.10 M = 0.00250

HNO2 + OH- >> NO2- + H2O

moles HNO2 = 0.00500 - 0.00250 = 0.00250
Moles NO2- = 0.00250

total volume = 0.075 L
concentration HNO2 = concentration NO2- = 0.00250 / 0.075 = 0.033 M

pKa = - log Ka =3.4

pH = 3.4 + log 0.033 / 0.033 = 3.4

3.

• 1000 mL x 0.05M = 50 millimols.
  x mL x 0.1M = ? mmols.
  -----------------
  ............HOCl + OH^- ==> OCl^- + H2O
  initial.....50......0........0........0
  add.................x................
  change......-x....-x.........x........x
  equil.......50-x....0.........x........x
  I used 3E-8 for Ka for HOCl
  pH = pKa + log(base)/(acid)
  8.00 = 7.52 + log(x)/(50-x)
  Solve for x = millimols OH^- added. Then mL = mmols/M. I get approximately 380 ml