Titration of KHP with NaOH: Finding the pH at various concentrations. Please help!?

KHP has only the Ka2 left from H2P

so there is only one PKa for KHP & it is 5.408,

which means its K = 3.91 e-6

2. What do you expect the pH to be at the midpoint of this titration?

pH at a mid point in a titration is = pKa = 5.408

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c) Calculate the pH at each of the following volumes of 0.20 M NaOH added to 20.00 mL of 0.20 M KHP.

adding NaOH:

0 mL,

k = [H][P] / [HP]

3.91 e-6 = [x][x] / [0.2]

x = [H] = 8.84e-4

pH = 3.05

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10 mL, 1/2 through the titration

mid point pH = pKa

pH = 4.93

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15 mL, you are 3/4 through

k = [H][P] / [HP]

3.91 e-6 = [x][0.15] / [0.05]

x = [H] =1.30e-6

pH = 5.88

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20 mL, you are done, but it is not pH = 7 when your acid has a K… there is hydrolysis , ( the P-2 is diluted to 1/2 strength)

P-2 in water –> HP- & OH-

K hydrolysis = K water / Ka = 1e-14 / 3.91e-6 = 2.56 e -9

2.56e-9 = [HP-] [OH-] / [P-2]

2.56e-9 = [X] [X] / [0.1]

2.56 e-10 = X squared

X = [OH-] = 1.60 e-5

p OH = 4.80

pH = 14 – 4.80 = 9.20

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22ml

has 2 mls of excess NaOH added

0.002 @ 0.20 mol/litre = 0.0004 moles OH in excess in 42 ml

0.0004 moles OH / 0.042 Litres = 0.009524 molar OH-

pOH = 2.02

pH = 11.98

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RE:

Titration of KHP with NaOH: Finding the pH at various concentrations. Please help!?

Answer the following questions for the titration of NaOH with KHP.

(pKa1(H2P) = 2.950, pKa2(H2P) =5.408).

1. What is the equivalence volume (Veq) when 20.00 mL of 0.20 M KHP is titrated with 0.20 M NaOH?

I got 20 mL for this one.

2. What do you expect the pH to be at the…

https://shorturl.im/awt55

So you’re trying to find the Molarity for NaOH? The equation for molarity is M=mols/L. You are given the mL and the M of KHP so you plug that into your equation to get .001 as your # of mols. From there simply substitute the volume of KHP for the volume of NaOH in order to get the concentration of .01 for NaOH. The equations would look like this. 0.050=mols/.02L(multiply 0.050 by .02)—>this gives you the # of moles which is .001. then you would do M=.001/.01((Liters of NaOH)—>this gives you the concentration of NaOH which is .01M Hope this helps!

Another method that would work would be to add a reagent to the NaOH that would precipitate all of the hydroxide ion, dry the precipitate thoroughly and then weigh the product. Perhaps Ferric chloride would work. This is called a gravimetric procedure. 3 NaOH(aq) + FeCl3(aq) ——-> 3 NaCl(aq) + Fe(OH)3(s)