A chemist has synthesized a monoprotic weak acid and wants to determine its Ka value. To do so, the chemist dissolves 2.00 mmol of the solid acid in 100.0 mL water and titrates the resulting solution with 0.0500 M NaOH. After 20.0 mL NaOH has been added, the pH is 6.00. What is the Ka value for the acid

1x10⁻⁶ is the Ka value for the acid

Explanation:

When a weak acid, HA, reacts with NaOH, some conjugate base, A⁻, is produced:

HA + NaOH → A⁻ + Na⁺ + H₂O

When HA is in excess, the NaOH added is equal to A⁻ produced and HA is Initial HA - NaOH added

After the reaction, the moles of HA are:

Initial moles HA: 2.00mmol

Moles NaOH: 0.020L * (0.050mol/L) = 1x10⁻³ moles = 1mmol

Moles HA: 2mmol - 1mmol = 1mmol

Moles A⁻ = Moles NaOH = 1mmol

Using H-H equation we can solve for pKa of the buffer:

pH = pKa + log [A⁻] / [HA]

Where [] could be taken as mmol of each species in the buffer:

6.00 = pKa + log 1mmol / 1mmol

6.00 = pKa

As pKa is -log Ka:

6.00 = -log Ka

10^-6 = Ka

i believe the answer is a.