A volume of 500.0 mL of 0.130 M NaOHNaOH is added to 625 mL of 0.200...

Question

Question

A volume of 500.0 mL of 0.130 M NaOHNaOH is added to 625 mL of 0.200 M weak acid (Ka=8.94×10−5). What is the pH of the resulting buffer?

HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)

science chemistry

Answer 1

Answer #1

We have , Molarity = No. of moles of solute / volume of solution in L

No. of moles of solute = Molarity volume of solution in L

No. of moles of NaOH = 0.130 mol / L 0.500 L = 0.0650 mol

No. of moles of HA = 0.200 mol / L 0.625 L = 0.125 mol

Consider reaction, HA + NaOH NaA + H₂O

Let's use Table.

Volume of solution = 500 + 625 = 1125 ml = 1.125 L

Solution contain weak acid HA & its salt NaA. This solution acts as a buffer solution & its pH is calculated by using Henderson's equation.

pH = pKa + log [ NaA ] / [ HA ]

pH = - log K a + log [ NaA ] / [ HA ]

pH = - log ( 8.94 10 ^-05 ) + log ( 0.0650 mol / 1.125 L ) / ( 0.0600 mol / 1.125 L )

pH = 4.05 + 0.0348

pH = 4.08

Answer 2

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