Question

Calculate the initial pH and the final pH after adding 0.010 mole of HCl to 500.0 mL of a buffer solution that is 0.125 M in CH_3COOH (K_a = 1.8 Times 10^-5) and 0.115 M CH_3COONa

Answer 1

pH = pka + log [salt]\[acid] {formula}

given: before adding HCL : concentration of salt CH₃COONa = 0.115 M , no.of moles = 0.05 mole

{ M= no. of moles*1000\ volume of solution in mL, no.of moles = (0.115*500)\ 1000= 0.05 mole }

concentration of acid CH₃COOH = 0.125 M, no. of moles= 0.06 mole

{ no.of moles= (0.125*500)\1000 = 0.06 mole}

K_a= 1.8*10^-5

   therefore, pka= -log K_a= - log (1.8*10^-5) = 4.74

by putting these values in above formula we get,

pH= 4.74 + log [0.115]\[0.125]

pH= 4.74+ log[0.92]

pH= 4.74+(-0.036)

pH= 4.704 -------------------------------------------------------------------------------initial pH

after adding HCL: no. of moles of HCL added = 0.01 mole

adding HCL increases the concentration of acid , decreases concentration of salt

because , H⁺ + CH₃COO^- --->  CH₃COOH

therefore, now concentration of salt CH₃COONa = 0.05-0.01= 0.04 mole

concentration of acid  CH₃COOH= 0.06+0.01= 0.07 mole

pH = pka + log [salt]\[acid] {formula}

pH= 4.74+ log [0.04]\[0.07]

pH = 4.74 + (-0.24)

pH= 4.49 --------------------------------------------------------------------final pH