Question #131928
Create a schematic diagram on how to prepare a Phosphate Buffer NaH2PO4 - K2HPO4 with a final volume of 500. mL, concentration of 0.10M with a pH of 7.20 from 0.20 M NaH2PO4 and solid K2HPO4.
Note: Ka = 6.3 x 10-8


Reminders: Show all calculations before preparing the schematic diagram.
For solid K2HPO4, calculate for the mass (based from calculated moles of salt) rather than the volume since this is a solid substance.
1
Expert's answer
2020-09-09T04:36:17-0400

H2PO4- acts as acid and HPO42- acts as salt.

pKa=log(Ka)pK_a = -log(K_a)

pKa=log(6.3×108)=7.2pK_a = -log(6.3\times 10^{-8}) = 7.2

The required pH = 7.20

pH=pKa+log[K2HPO4][NaH2PO4]pH = pK_a + log\frac{[K_2HPO_4]}{[NaH_2PO_4]}

Since pH = pKa the salt and acid concentrations are equal.

[NaH2PO4] = [K2HPO4]

Given the total concentration = 0.10 M

Therefore the concentration of [NaH2PO4] = 0.05 M

The stock solution of NaH2PO4 = 0.2 M

Therefore the volume of stock solution required V=500×0.050.2=125  mLV = 500 \times \frac{0.05 }{ 0.2} = 125 \;mL

We need 125 mL of 0.2 M NaH2PO4 solution.

The concentration of [K2HPO4] = 0.05 M

The total volume of the buffer solution V = 500 mL

Therefore moles of K2HPO4 in the buffer

n=500  mL×0.05  mol1000  mL=0.025  molesn = 500\; mL \times \frac{0.05\; mol }{ 1000 \;mL} = 0.025 \;moles

M(K2HPO4) = 174 g/mol

m=n×M

m(K2HPO4)=0.025  mol×174  g/mol=4.35  gm(K_2HPO_4) = 0.025\; mol \times 174\;g/mol = 4.35 \;g


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