Calculate the pH of the titration of 100.00 mL of 0.200 M HOCI with 0.400 M KOH after 0, 25.00, 50.00, & 75.00 mL of KOH have been added. The Ka for HOCI is .000000035...Please Help!!!!
First write the equation and balance it.
KOH + HOCl ==> KOCl + H2O
moles HOCl = M x L = ??
To do these you simply need to figure out what you have in the solution, then apply the appropriate chemical properties and calculate. I will help you get started.
The second thing you need to do is to calculate the mL for the equivalence point.
mL x M = mL x M and I get 50.00 mL KOH.
a. 0 mL of 0.4 M KOH. Here you have 100 mL of 0.2 M HOCl. You have Ka, calculate H^+ and pH.
b. 25 mL. At 25 mL you aren't to the equivalence point; therefore, you have a mixture of HOCl and KOCl. Use the buffer equation
c. 50.0 mL. You are at the equivalence point; therefore, the pH will be determine by the hydrolysis of the salt. Calculate concn of salt and hydrolyze it (NaOCl).
OCl^- + HOH ==>HOCl + OH^-
Set up ICE chart and substitute into the Kb expression
Kb = (Kw/Ka) = (HOCl)(OH^-)/(OCl^-), solve for OH^-, convert to pOH, then to pH.
d. 75 mL. This is past the equivalence point (by 25 mL), therefore, you have straight OH^- from the KOH (but diluted) so (OH^-) = 25 mL x 0.4M moles and that is in a volume of 175 mL.
Post your work and explain what you don't understand if you get stuck.
To calculate the pH of the titration, we need to determine the concentration of the resulting solution at each addition of KOH. We can use the balanced chemical equation and stoichiometry to do this.
The balanced chemical equation for the reaction between HOCI (hypochlorous acid) and KOH (potassium hydroxide) is:
HOCI + KOH -> KCI + H2O
Based on the balanced equation, we can see that the reaction between HOCI and KOH is a 1:1 ratio. This means that for every 1 mole of HOCI, we'll need 1 mole of KOH.
Given:
Initial volume of HOCI (V1) = 100.00 mL = 0.10000 L
Concentration of HOCI (C1) = 0.200 M
Let's calculate the moles of HOCI (n1) initially present:
n1 = C1 * V1
= 0.200 M * 0.10000 L
= 0.0200 moles
Now, let's calculate the moles of KOH (n2) added at each point:
For 0 mL KOH added (before any addition):
n2 = 0 moles
For 25.00 mL KOH added:
V2 = 0.02500 L
C2 = 0.400 M
n2 = C2 * V2
Similarly, for 50.00 mL and 75.00 mL KOH added, we can calculate n2 using the corresponding volumes and concentrations.
Next, let's calculate the moles of HOCI remaining at each point:
n1' = n1 - n2
Finally, let's calculate the concentration of HOCI at each point:
V' = V1 + V2
C' = n1' / V'
Now, we can calculate the pH at each point using the Henderson-Hasselbalch equation:
pH = -log10(sqrt(Ka / C'))
Where Ka is the acid dissociation constant for HOCI.
Using this process, you can calculate the pH of the titration at each point with the provided information and the equations mentioned above.