A 1 L buffer solution contains 0.3 M acetic acid and 0.3 M sodium acetate. A certain mass of solid NaOH is added to the solution, which changes its pH by 0.30 units. Assuming no change in the volume of the solution, the mass of solid NaOH added is: (Given: pK _ a of acetic acid = 4.74 , 2 = 0.30 , Molar mass of NaOH = 40 g mol ⁻¹ )

Step-by-step solution

Initial concentration of acetic acid = 0.3 M and sodium acetate = 0.3 M . Initial pH = pK _ a + ( 0.3 0.3 ) = 4.74 . Addition of a strong base ( NaOH ) will increase the pH . Thus, the new pH is: pH _ new = 4.74 + 0.30 = 5.04 Let x be the number of moles of solid NaOH added to 1 L of the buffer. The added NaOH reacts with acetic acid to form more sodium acetate. New moles of salt = 0.3 + x New moles of acid = 0.3 - x Applying the Henderson-Hasselbalch equation: 5.04 = 4.74 + ( 0.3 + x 0.3 - x ) 0.30 = ( 0.3 + x 0.3