Question

What volume (to the nearest 0.1 mL) of 5.30-M NaOH must be added to 0.700 L of 0.350-M HNO2 to prepare a pH = 3.70 buffer?

What volume (to the nearest 0.1 mL) of 4.80-M HCl must be added to 0.550 L of 0.250-M K2HPO4 to prepare a pH = 8.00 buffer?

Answer 1

1)

30.8ml

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

where,

[A^-] = conjucate base ( NO2^- is conjucate base)

[HA] = weak acid ( HNO2 is weak acid)

required pH = 3.70

pKa of HNO2 = 3.398

substituting the values

3.70 = 3.398 + log([NO2^-]/[HNO2])

log([NO2^-]/[HNO2]) = 0.302

[NO2^-]/[HNO2] = 2.004

moles of NO2^-/moles of HNO2 = 2.004

moles of NO2^- = moles of HNO2 × 2.004

Total buffer concentration = 0.350M

Total number of moles = (0.350mol/1L)×0.700L = 0.2450mol

so,

moles of HNO2 × 2.004moles of HNO2 = 0.2450mol

3.004moles of HNO2 = 0.2450mol

moles of HNO2 = 0.2450mol/3.004 = 0.08156 mol

moles of NO2^- = 0.2450 mol - 0.08156mol = 0.1634mol

reaction between NaOH and HNO2 is 1:1 reaction

HNO2 + OH^- -------> NO2^- + H2O

So,

0.1634moles of OH^- react with 0.1634moles of HNO2 to produce 0.1634moles of NO2^-

Therefore,

No of moles of NaOH required = 0.1634

Volume of 5.30M NaOH solution containing 0.1634moles of NaOH = (1000ml/5.30mol)×0.1634mol = 30.8ml

2)

3.9ml

Explanation

pH = pKa + log([A^-]/[HA])

pKa of KH2PO4 = 7.20

required pH = 8.00

conjucate base is HPO4^2-

weak acid is H2PO4^-

8.00 = 7.20 + log([HPO4^2-]/[H2PO4^-])

log([HPO4^2-]/[H2PO4^-]) = 0.80

[HPO4^2-]/[H2PO4^-] = 6.31

[HPO4^-] = [H2PO4^-] ×6.31

moles of HPO4^- = moles of H2PO4^- × 6.31

Total moles of buffer = (0.250moles /1000ml)× 550ml = 0.1375mol

moles of H2PO4^- + 6.31 moles of H2PO4^- = 0.1375mol

7.31moles of H2PO4^- = 0.1375mol

moles of H2PO4^- = 0.01881mol

moles of HPO4^2- = 0.1375mol - 0.01881mol = 0.1187

reaction between HPO4^2- and HCl is 1:1 molar reaction

H⁺ + HPO4^2- ------> H2PO4^-

0.01881moles of HCl react with 0.01881moles of HPO4^2- to produce 0.01881moles of H2PO4^-

No of moles of HCl required = 0.01881mol

Volume of 4.80M HCl solution containing 0.01881 moles of HCl = (1000ml/4.80mol)×0.02179mol = 3.9ml