Calculate the ph for each of the following cases in the titration of 25.0 ml of 0.150 m sodium hydroxide with 0.150 m hbr(aq). (1 points each) (a) before addition of any hbr (b) after addition of 5.0 ml of hbr (c) after addition of 15.0 ml of hbr (d) after addition of 25.0 ml of hbr (e) after addition of 40.0 ml of hbr (f) after addition of 60 ml of hbr

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

e) pH = 1.46

f) pH = 1.21

Explanation:

HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

⇒ C NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

⇒ pOH = - Log ( 0.150 )

⇒ pOH = 0.824

⇒ pH = 14 - pOH

⇒ pH = 13.176

b) after addition 5mL HBr:

⇒ C NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

⇒ C HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M

⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ C NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ C HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

⇒ pH = 7.0

d) after addition 40mL HBr:

⇒ C HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

⇒ [ H3O+ ] = 0.035 M

⇒ pH = 1.46

d) after addition 60mL HBr:

⇒ C HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M

⇒ [ H3O+ ] = 0.062 M

⇒ pH = 1.21

a) 13.18

b) 13.0

c) 12.57

d) 7.00

e) 1.46

f) 1.21

Explanation:

In the titration, the acid (HBr) will react with the base (NaOH) in a neutralization reaction, to form salt and water. The reaction happens in the stoichiometry 1:1, so the reactant with fewer moles will be consumed.

Knowing that pH = 14 - pOH, and pOH = -log[OH-], and also pH = -log[H+], let's analyze the itens:

a) First there is only NaOH, so, it is dissociated and the [OH-] = [NaOH] = 0.150 M

pOH = -log(0.150) = 0.82

pH = 14 -0.82 = 13.18

b) When 5.0 mL (0.005 L) of HBr are added, it's added

nHBr = 0.005 L* 0.150 M = 0.00075 mol

So, 0.00075 mol of H+ reacts with OH-. The initial number of moles of OH- was

n = 0.025 L * 0.150 M = 0.00375 mol

After the reaction it will be:

n = 0.00375 - 0.00075 = 0.003 mol

The final volume is the sum of the volume of the two substances (30 mL = 0.03 L), so:

[OH-] = moles/volume = 0.003/0.03 = 0.1 M

pOH = -log(0.1) = 1

pH = 14 - 1 = 13

c) Doing the same thing as letter "b":

nHBr = 0.015L * 0.150 M = 0.00225 mol

nOH- = 0.00375 - 0.00225 = 0.0015 mol

Volume = 25 mL + 15 mL = 40 mL = 0.04 L

[OH-] = 0.0015/0.04 = 0.0375 M

pOH = -log(0.0375) = 1.43

pH = 14 - 1.43 = 12.57

d) At this point occurs the total neutralization, because the number of moles of the acid is equal to the number of moles of the base. Thus, pH = 7.00.

e) Know, the acid will be in excess.

nHBr = 0.04L* 0.150M = 0.006 mol

nH+ = 0.006 - 0.00375 = 0.00225 mol

Volume = 40 mL + 25 mL = 65 mL = 0.065 L

[H+] = 0.00225/0.065 = 0.035 M

pH = -log(0.035) = 1.46

f) Doing the same thing as letter "e"

nHBr = 0.06L * 0.150 M = 0.009 mol

nH+ = 0.009 - 0.00375 = 0.00525 mol

Volume = 60 mL + 25 mL = 85 mL = 0.085 L

[H+] = 0.00525/0.085 = 0.062

pH = -log(0.062) = 1.21

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