To prepare a buffer solution of pH = 5.2 , a student mixes 100 mL of 0.1 M weak acid HA with V mL of 0.1 M NaOH solution. The value of V is (Given: K_a( HA ) = 2.5 10⁻⁵ , 2 = 0.3 , 2.5 = 0.4 )

Step-by-step solution

First, calculate the p K_a of the weak acid HA : p K_a = - (K_a) = - (2.5 10⁻⁵) = 5 - 2.5 = 5 - 0.4 = 4.6 Using the Henderson-Hasselbalch equation for an acidic buffer: pH = p K_a + ( [ Salt ] [ Acid ] ) 5.2 = 4.6 + ( [ Salt ] [ Acid ] ) ( [ Salt ] [ Acid ] ) = 0.6 Since 4 = 2 2 = 2 0.3 = 0.6 , we have: [ Salt ] [ Acid ] = 4 Now, consider the neutralization reaction between HA and NaOH : Initial millimoles of HA = 100 mL 0.1 M = 10 mmol Millimoles of NaOH added = V mL 0.1 M = 0.1V mmol Since NaOH is the limiting re