Respuesta :
Answer:
Initial pH of this pH = 7.453
pH after addition of 150.0 mg of HBr = 7.35
pH after addition of 85.0 mg of NaOH 0.154 = 7.56
Explanation:
Since Ka value isn't given
so we use Ka value of HClO (hypo chlorous acid) = 3 x 10⁻⁸
pKa = - logKa = 7.52
Part A
Using Henderson equation
pH = pka + log[tex]\frac{[Conjugate base]}{[Acid]}[/tex]
pH = 7.52 + log [tex]\frac{0.15}{0.175}[/tex]
pH = 7.453
Part B
pH after addition of 150 mg of HBr
moles of HBr
[tex]\frac{Mass}{Molar mass} \\= \frac{150 X10^{-3}g }{80.91} \\= 0.00185 } mole[/tex]
Moles of NaOCl in 100 ml buffer solution = [tex]\frac{0.15X100}{1000}[/tex] = 0.015
Moles of HClO in 100 ml buffer solution = [tex]\frac{0.175X100}{1000}[/tex] = 0.0175
Since H⁺ concentration furnished by HBr acid make a common ion effect . So the following reaction carried out
ClO⁻ + H⁺ → HClO
So the remaining concentration of ClO⁻ in solution = 0.015 - 0.000185
= 0.0132
moles of HClO = 0.0175 + 0.00185
= 0.0194
Using Henderson equation pH = Pka + log[tex]\frac{Conjugate base}{Acid}[/tex]
= 7.52 + log[tex]\frac{0.0132}{0.0194}[/tex]
= 7.35
Part C
pH after addition of 85 mg of HBr
moles of NaOH
[tex]\frac{Mass}{Molar mass} \\= \frac{85 X10^{-3}g }{40} \\= 0.00213 } mole[/tex]
So the remaining concentration of ClO⁻ in solution = 0.015 + 0.00213
= 0.0171
Moles of concentration of ClO⁻ = 0.171(M)
moles of HClO = 0.0175 - 0.00213
= 0.0154
Moles in 100 ml Buffer = 0.154(M)
Using Henderson equation pH = Pka + log[tex]\frac{Conjugate base}{Acid}[/tex]
= 7.52 + log[tex]\frac{0.171}{0.154}[/tex]
= 7.56
