Question

What masses of bromoacetic acid (CH2BrCOOH) and sodium bromoacetate (CH2BrCOONa) are needed to prepare 1.00 L of pH = 2.500 buffer if the total concentration of the two components is 0.200 M?

Answer 1

ans)

from above data that

given that

pH = 2.5

Ka of bromoacetic acid is =2.0 × 10–3.

so the pKa = 2.70

total moles of buffer = 1 x 0.200 = 0.200

pH = pKa + log [salt / acid]

2.5= 2.70 + log [CH2BrCOONa /CH2BrCOOH ]

[CH2BrCOONa /CH2BrCOOH ] = 0.6309

[CH2BrCOONa + CH2BrCOOH ] = 0.200

0.6309 CH2BrCOOH + CH2BrCOOH = 0.200

moles of CH2BrCOOH = 0.1226 mol

substitue this value in above equation

moles of CH2BrCOONa = 0.0774 mol

mass of CH2BrCOOH =moles*molar mass

= 0.1226 mol*138.95 g/mol

=17.03 g

mass of CH2BrCOONa =moles*molar mass

=0.0774 mol*160.93 g/mol

=12.45 g