Respuesta :
From the equation kb = 1.8 x 10^-5
Therefore;
pKb = - log 1.8 x 10^-5 = 4.7
Moles NH3 in 75 ml = (75/1000) L x 0.200 M=0.0150
Moles HNO3 in 19 ml = (19/1000) L x 0.500 M= 0.0095
The net reaction is
NH3 + H+ = NH4+
Moles NH3 in excess = 0.0150 - 0.0095 =0.0055
Moles NH4+ formed = 0.0095
Total volume = 75.0 + 19.0 = 94.0 mL = 0.094 L
[NH3]= 0.0055/ 0.094 L=0.0585 M
[NH4+] = 0.0095/ 0.094 L = 0.1011M
pOH = pKb + log [NH4+]/ [NH3]
= 4.7 + log 0.1011/ 0.0585
= 4.938
pH = 14 - pOH
= 14 – 4.94
=9.06
pH after the addition of 19.0 ml of HNO₃ : 9.018
Further explanation
The pH value of a reaction between strong acid HNO₃ and weak base NH₃ can be estimated from the rest of the reaction product
1. If the remainder of the reaction results obtained the remaining strong acid HNO₃, the pH is sought from the concentration of [H⁺] using the formula
[H⁺] = a. M
a = valence of acid / amount of H⁺ released
M = acid concentration
2. When strong acids and weak bases react, they form salts that are acidic, calculating the pH using the pH hydrolysis formula
[tex]\displaystyle [H +]=\sqrt{\frac{Kw}{Kb}.M }[/tex]
where
M = concentration of salt anion
3. If the remainder of the reaction results are a weak base remaining and the salt, the solution will form a base buffer solution and search for pH using the formula base buffer pH
[tex]\displaystyle [OH-]=Kb\times\frac{weak\:base\:mole}{salt\:mole\times valence}[/tex]
We count the moles of each reactant:
NH₃ mole = 75 ml x 0.2 M = 15 mlmol
mole HNO₃ = 19 ml x 0.5 M = 9.5 mlmol
NH₃ + HNO₃ ---> NH₄NO₃
15 9.5
9.5 9.5 9.5
5.5 0 9.5
so there are remaining weak bases NH₃ = 15 - 9.5 = 5.5 mmol
Then a buffer solution is formed
[OH⁻] = Kb x [weak base mole] / [salt mole x valence]
[tex]\displaystyle [OH-]=1.8.10^{-5}\times\frac{5.5}{9.5\times 1}\\\\pOH=-log\:1.042.10^{-5}\\\\pOH=4.982\\\\pH=14-pOH\\\\pH=9.018[/tex]
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Keywords : pH, acid, base, HNO₃,NH₃
