How many mL of 0.80 molar NaOH must be added to 20 ml of 0.40 of molar propionic acid, in order to produce a solution with a pH of 4.89? (PKa of propionic acid is 4.89.)
To solve this question, we need to set up a balanced equation for the reaction between NaOH (sodium hydroxide) and propionic acid (C2H5COOH):
C2H5COOH + NaOH -> C2H5COONa + H2O
Taking into account the stoichiometry of the equation, we can determine the number of moles of propionic acid present in 20 mL of a 0.40 M solution.
Using the formula:
Moles = Volume (in liters) x Concentration (in Molarity)
Moles of propionic acid = 20 mL x (1 L / 1000 mL) x 0.40 M
= 0.008 moles
Since the pKa of propionic acid is given as 4.89, we know that when one-half of the acid is neutralized by the base, the pH of the resulting solution will be equal to the pKa.
Therefore, we need to find the number of moles of NaOH required to neutralize half of the propionic acid:
Moles of NaOH = 0.008 moles / 2
= 0.004 moles
To find the volume of 0.80 M NaOH needed to obtain 0.004 moles, we can rearrange the moles equation to:
Volume (in liters) = Moles / Concentration (in Molarity)
Volume of NaOH = 0.004 moles / 0.80 M
= 0.005 L
Since the volume is currently in liters, we need to convert it to milliliters:
Volume of NaOH = 0.005 L x (1000 mL / 1 L)
= 5 mL
Therefore, 5 mL of 0.80 M NaOH must be added to 20 mL of 0.40 M propionic acid to produce a solution with a pH of 4.89.