A student titrates 0.100M KOH, potassium hydroxide, into 50.0ml of 0.10M HCOOH, formic acid (Ka=1.8 x 10^-4. What is the pH of the mixture after 25.0ml of the KOH has been added
moles KOH initially = M x L = ??
moles HCOOH initially = M x L = ??
KOH + HCOOH ==> HCOOK + H2O
Use the above information on the first two lines to set up an ICE chart. then use the Henderson-Hasselbalch equation to solve for pH of the buffer solution that has been prepared.
The other way to do this is to realize that 25.0 mL is just half-way to the equivalence point which means pH = pKa.
To determine the pH of the mixture after 25.0 mL of 0.100 M KOH has been added, we need to understand the acid-base reaction between KOH (a strong base) and HCOOH (a weak acid).
First, let's calculate the initial moles of HCOOH present in the 50.0 mL solution before titration.
Molarity (M) = Moles (mol) / Volume (L)
0.10 M = Moles / 0.050 L
Moles = 0.10 M x 0.050 L = 0.005 mol
Now, let's determine the moles of HCOOH that have reacted with the added 25.0 mL of KOH.
Molarity (M) = Moles (mol) / Volume (L)
0.100 M = Moles / 0.025 L
Moles = 0.100 M x 0.025 L = 0.0025 mol
Since the reaction between KOH and HCOOH is a 1:1 ratio, 0.0025 mol of HCOOH has reacted. Subtracting this from the initial moles of HCOOH gives us the moles remaining.
Moles remaining = 0.005 mol - 0.0025 mol = 0.0025 mol
Now, let's calculate the concentration of formate ion (HCOO-) formed from the reaction.
Molarity (M) = Moles (mol) / Volume (L)
M = 0.0025 mol / 0.075 L (0.050 L + 0.025 L)
M = 0.033 M
With this information, we can calculate the pOH of the mixture.
pOH = -log10[M(OH-)]
pOH = -log10[0.033]
Using a calculator, pOH = 1.48
Finally, we can find the pH of the mixture using the equation:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.48
Using a calculator, pH = 12.52
Therefore, the pH of the mixture after adding 25.0 mL of 0.100 M KOH is approximately 12.52.