A chemist synthesized a monoprotic weak acid, (HA) and 2.00 mmol of the solid acid is...

A chemist synthesized a monoprotic weak acid, (HA) and 2.00 mmol of the solid acid is dissolved in 100.0 ml of water. The acid is titrated with 20.0 ml of 0.0500 M NaOH. If the final solution has a pH of 6.00, what's the Ka of the acid?

Expert Solution

PH = 6.00

weak acid = 2.0 m mol= 2.0x10^-3 = 0.002 mole

NaOH = 20.0ml 0f 0.0500M

Number of moles of NaOH = 0.05x0.02 L =0.001 mole

HA + NaOH --------------- NaA + H2O

0.002 mol 0.001 mole 0 0

0.002-.0001 0.001-0.001 0.001

0.001 0 0.001

Number of moles of HA = 0.001 mole

Number of moles of NaA = 0.001 moles

PH= PKa + log[salt/acid]

6.00 = PKa + log[0.001/0.001]

6.00 = PKa + 0

PKa = 6.0

-log[ka] = 6.0

Ka = 10^-6.0

Ka = 10^-6

queen_honey_blossom answered 2 years ago

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