Respuesta :
Answer:
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:
- NaOH → Na+ + OH-
⇒ 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
Answer:
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
