Question

Making a buffer soln with K2HPO4 and KH2PO4?

1. How many grams of acid would I need to make .08M soln in 100ml? And how many grams of base would I need to make a .05M soln in 100ml?

2. What is the theoretical buffer pH? What is the theroetical buffer pH after lmL of .1M NaOH is added?

Answer 1

Solution :-

Part 1)

K2HPO4 is the base and KH2PO4 is the acid

Volume of buffer = 100 ml

Molarity of acid =0.08 M

Molarity of base =0.05 M

Using the molarity and volume we can find the moles of each

Moles = molarityx volume in liter

Moles of acid (KH2PO4)= 0.08 molper L * 0.100 L= 0.008 mol

Moles of base (K2HPO4) = 0.05 mol per L *0.100 L = 0.005 mol

Now lets calculate the mass of each

Mass = moles x molar mass

Mass of acid (KH2PO4) = 0.008 mol * 136.09 g per mol =1.09 g

Mass of base (K2HPO4)= 0.005 mol * 174.18 g per mol = 0.871 g

Therefore we need 1.09 g acid (KH2PO4) and 0.871 g base (K2HPO4)

Part 2)

Calculating the theoretical pH of the buffer

pH= pka + log [base]/[acid]

pka of the acid H2PO4^- = 7.21

lets put the values in the formula and calculate the pH

pH= 7.21+ log [0.05]/[0.08]

pH= 7.01

therefore the theoretical pH of the buffer is 7.01

now lets calculate the pH after adding 1ml of 0.1 M NaOH

moles of NaOH = 0.1 mol per L * 0.001 L = 0.0001 mol

NaOH reacts with acid therefore the moles of acid will decrease and moles of base will increase as the number of moles of NaOH added

after the reaction moles of acid = 0.008 mol – 0.0001 mol= 0.0079 mol

moles of base = 0.005mol + 0.0001mol = 0.0051 mol

total volume = 100 ml + 1 ml = 101 ml = 0.101 L

new molarity of acid = 0.0079 mol /0.101 L= 0.07822 M

new molarity of base = 0.0051 mol / 0.101 L = 0.0505 M

pH= pka + log [base]/[acid]

pH= 7.21+ log [0.0505]/[0.07822]

pH= 7.02