Question

how many grams of NaOH must be added to 1L of a 0.1 M KH2PO4 to prepare a pH 7.42 buffer. answer given by professor is 2.47 I need help understanding how to solve

Answer 1

we know that

moles = concentration x volume

so

moles of H2P04- = 0.1 x 1 = 0.1

now

let y moles of NaOH is added

the reaction is

H2P04- + OH- --> HP042- + H20

we can see that

moles of H2P04- reacted = moles of OH- added = y

moles of H2P04- left = 0.1 - y

also

moles of HP042- formed = moles of OH- added = y

now

we know that

the buffer combination is H2P04- and HP042-

now

for buffers

pH = pKa + log [ conjugate base / acid ]

so

pH = pKa + log [ HP042- / H2P04- ]

we know that

pKa for H2P04- is 7.2

given

pH = 7.42

so

7.42 = 7.2 + log [ HP042- / H2P04- ]

[ HP042- / H2P04- ] = 1.66

now

we know that

moles = concentration x volume

as final volume is same for both

ratio of concentration = ratio of moles

so

moles of HP042- / moles of H2P04- = 1.66

y / ( 0.1 - y) = 1.66

y = 0.166 - 1.66 y

y = 0.0624

so

moles of NaOH added = y = 0.0624

now

mass = moles x molar mass

so

mass of NaOH = 0.0624 x 40

mass of NaOH = 2.496

so

2.496 grams of NaOH must be added