Question

a 30.0 ml sample of 0.75 M hypociouous acid, hocl, is titrated with 1.25 M KOH....

a 30.0 ml sample of 0.75 M hypociouous acid, hocl, is titrated with 1.25 M KOH. what is the ph of the solution after 7.0 ml of KOH is added

Homework Answers

Answer #1

[HOCl] = molarity x volume in Litres = 0.75 M x 0.03 L = 0.0225 mol

[KOH] = molarity x volume in Litres = 1.25 M x 0.007 L = 0.00875 mol

  HOCl +   KOH -----------------> KOCl + H2O

0.0225 mol 0.00875 mol   

-------------------------------------------------------------------------------------------------------------

0.0225 -0.00875 0 0.00875 mol

= 0.01375 mol

Hence,

[HOCl] = 0.01375 mol

[KOCl] =  0.00875 mol

We know that pKa of HOCl = 7.53

pH = pKa + log [salt] / [acid]

= pKa + log [KOCl] / [HOCl]

= 7.53 + log (0.00875/0.01375)

= 7.33

pH = 7.33

Therefore,

pH of the given solution = 7.33

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A 30.0 mL sample of 0.05 M HCLO is titrated by 0.0250 M KOH solution. Ka...
A 30.0 mL sample of 0.05 M HCLO is titrated by 0.0250 M KOH solution. Ka for HCLO is 3.5 x 10-8. Calculate the pH when no base has been added, the pH when 30.0 mL of the base has been added ,. the pH at the equivalence point and the pH when an additional 4.00 mL of the KOH solution has been added beyond the equivalence point
A 30.0 mL sample of 0.05 M HCLO is titrated by 0.0250 M KOH solution. Ka...
A 30.0 mL sample of 0.05 M HCLO is titrated by 0.0250 M KOH solution. Ka for HCLO is 3.5 x 10-8. Calculate..... a. the pH when no base has been added b. the pH when 30.0 mL of the base has been added c. the pH at the equivalence point d. the pH when an additional 4.00 mL of the KOH solution has been added beyond the equivalence point
1) 30.0 mL of 0.44M hydrazoic acid; (HN3; Ka=2.6 x 10^-5) is titrated with a 0.22...
1) 30.0 mL of 0.44M hydrazoic acid; (HN3; Ka=2.6 x 10^-5) is titrated with a 0.22 M KOH solutions . What's pH of solution after adding 60.0 mL of base? 2) 25.0mL of 0.800M HClO4 is titrated with a 0.300M KOH solution. What is the H3O+ concentration after 10.0 mL of KOH are added?
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a...
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a 0.144 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. (The Ka for acetic acid is 1.76 x 10^-5). 10.0 mL of base = 20.0 mL of base = 30.0 mL of base =
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a...
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a 0.144 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. (The Ka for acetic acid is 1.76 x 10^-5). 10.0 mL of base = 20.0 mL of base = 30.0 mL of base =
A 25.0-mL sample of a 0.350 M solution of aqueous trimelhylamine is titrated with a 0.438...
A 25.0-mL sample of a 0.350 M solution of aqueous trimelhylamine is titrated with a 0.438 M solution of HCI. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pK_b of (CH_3)_3N = 4.19 at 25 degree C. pH after 10.0 mL of acid have been added: pH after 20.0 mL of acid have been added: pH after 30.0 mL of acid have been added:
16.5 mL of 0.168 M diprotic acid (H2A) was titrated with 0.118 M KOH. The acid...
16.5 mL of 0.168 M diprotic acid (H2A) was titrated with 0.118 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the first equivalence point occur? 20.8 mL of 0.146 M diprotic acid (H2A) was titrated with 0.14 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the second equivalence point occur? Consider the titration of...
A 25.0 mL sample of solution of unkown acid, HX, is titrated with 2.87 M KOH....
A 25.0 mL sample of solution of unkown acid, HX, is titrated with 2.87 M KOH. If 59.9 mL of KOH was required to neutralize the sample, find the molaritu of fhe original HX solution.
A 25.0 mL sample of a 0.0500 M solution of aqueous trimethylamine is titrated with a...
A 25.0 mL sample of a 0.0500 M solution of aqueous trimethylamine is titrated with a 0.0625 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25°C.
A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the...
A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 95 mL of KOH. The Ka of HF is 3.5 × 10-4.