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1) 25.00 mL of .200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
2) What is the pH at the equivalence point when 50.00 mL of a .150 M solution of acetic acid (CH3COOH) is titrated with .10 M NaOH to its end point?
3) A 25.0 mL sample of a .290 M solution of aqueous trimethylamine is titrated with a .363 M solution of HCl. Calculate the pH of the solution
a) after 10.0 mL of acid have been added:
b) after 20.0 mL of acid have been added:
c) after 30.0 mL of acid have been added:
*pKb of (CH3)3N=4.19 at 25 degrees C.11
4) Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.00 mL sample of the cleaner, the equivalence point is reached after 16.81 mL of .135 M HCl has been added. What is the
a) initial concentration of ammonia in the solution?
b) pH of the solution at the equivalence point?



Answer :

AnjaliP102

The pH of water indicates how acidic or basic it is.

How to find pH value?

1. The moles of equivalence of sodium hydroxide and nitrous acid are as follows:

nNaOH = nHNO₃

             = 25.00 x .200

             = .005 mol

The added base has a volume of 5.00 mL.

Nitrogen dioxide has a mole of 0.005 and a volume of 30 mL at the equivalence point.

Nitrogen dioxide's molar concentration will be 0.167 M.

The base dissociation constant is calculated using the ICE table as follows:

Kw = Ka x Kb

Kb = 1 x 10⁻¹⁴ / 4.6 x 10⁻¹⁴

= 2.174 x 10⁻¹¹

The base's ionization constant is given as,

Kb = [ HNO₂] [ OH⁻] / [NO₂⁻]

2.174 X 10⁻¹¹ = x² / 0.167 - x

                    = 1.905 x 10⁻6

pOH is now calculated as follows:

pOH = -log[OH⁻]

         = -log(1.905 x 10⁻⁶)

          = 5.72

The titration's pH is calculated as follows:

pH + pOH = 14

pH = 14 - 5.72

= 8.27

As a result, the pH is 8.27.

2. Given,

The acetic acid volume = 50.0 mL

The molarity of acetic acid = 0.150 M

The molarity of NaOH = 0.10 M

HA + H2O ⇆ A- + H3O+

Initial concentration is :

[HA] = 0.150 M

[A-] = 0M

[H3O+] = 0M

Concentration at equilibrium

[HA] = 0.150 - x M

[A-] = xM

[H3O+] = xM

Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5

0.150 >>> x

x²/(0.150) = 1.76*10^-5

x² =0.00000264

x = 0.00162

[H3O+] = 0.00162 M

pH = -log [H3O+] = -log (0.00162) = 2.79

3. a) by adding 10 mL of HCl

moles of (CH3)3N = molarity * volume

                                                               = 0.29 m * 0.025 L

                                                                = 0.00725M moles

moles of HCl = molarity * volume

                                                         = 0.3625 m * 0.01L

                                                         = 0.003625 moles

moles of (CH3)3N remaining    = moles of (CH3)3N - moles of HCl

                                                        = 0.00725 - 0.003625

                                                        = 0.003625 moles

the total volume = 0.01 L + 0.025L = 0.035 L

[(CH3)3N] = moles remaining / total volume

                  = 0.003625 moles / 0.035L

                   = 0.104 M

pKb = - logKb

4.19 = -logKb

Kb = 6.5 x 10^-5

Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]

by using ICE table :

[(CH3)3NH+] = X & [OH] = X

[(CH3)3N] = 0.104 -X

By substituting:

6.5 x 10^-5 = X^2 / (0.104-X)  by solving for X

X = 0.00257 M

[OH-] = X = 0.00257 M

POH = -log[OH]

          = -log0.00257

          = 2.5

PH = 14 - POH

        = 14 - 2.5

        = 11.5

b) by adding 20ML of HCL:

The moles of HCl = molarity * volume

                      = 0.3625 m * 0.02 L

                      = 0.00725 moles

The neutralization  of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:

When total volume = 0.02L + 0.025 = 0.045L

∴ [ (CH3)3NH+] = moles / total volume

                         = 0.003625 / 0.045L

                         = 0.08 M

When Ka = Kw / Kb

Here, Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14

By substitution:

Ka = (1 x 10^-14) / (6.5 x 10^-5)

   = 1.5 x 10^-10

Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]

By substituting:

1.5 x 10^-10 = X^2 / (0.08 - X)  solving X

X = 3.5 x 10^-6  M

[H+]= X = 3.5 x 10^-6 M

The PH = -log[H+]

       = -log(3.5 x 10^-6)

      = 5.5

C) by adding 30ML of HCl:

The no : of moles HCl = molarity * volume

                      = 0.3625m * 0.03L

                      = 0.011 moles

The moles of (CH3)3N neutralized = 0.003625 moles

The no : of moles of HCl remaining    = moles HCl - moles (CH3)3N neutralized

                                               = 0.011moles - 0.003625moles

                                               = 0.007375 moles

Therefore total volume = 0.025L + 0.03L

                               = 0.055L

The [H+] = moles / total volume

          = 0.007375 mol / 0.055L

           = 0.134 M

Therefore the PH = -log[H+]

       = -log 0.134

       = 0.87

To learn more about pH value refer to :

https://brainly.com/question/26424076

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