Respuesta :
Answer:
pH of buffer before addition of Strong acid = 4.71
pH of buffer after addition of Strong acid = 2
Explanation:
According to Henderson's equation
⇒pH = pKa +㏒[tex]\frac{[salt]}{[Acid]}[/tex]
⇒pH = 4.74 + log([tex]\frac{0.115}{0.12}[/tex])
⇒pH = 4.74 - 0.03
⇒pH = 4.71
⇒[H⁺] = 1.9 x 10⁻⁵
pH of this buffer after 0.01 mole of strong acid addition
⇒pH = - log[H⁺]
⇒pH = - log(1.9 x 10⁻⁵ + 0.01)
⇒pH = 2
pH of buffer before addition of Strong acid = 4.71
pH of buffer after addition of Strong acid = 2
According to Henderson's equation:
pH = pKa + log ([salt]/ [acid])
pH = 4.74 + log (0.115/0.12)
pH = 4.74 - 0.03
pH = 4.71
[H⁺] = 1.9 x 10⁻⁵
pH of this buffer after 0.01 mole of strong acid addition:
pH = - log[H⁺]
pH = - log(1.9 x 10⁻⁵ + 0.01)
pH = 2
Thus,
pH of buffer before addition of Strong acid = 4.71
pH of buffer after addition of Strong acid = 2
Find more information about pH here:
brainly.com/question/13557815
