Question

Acid Ka H2PO4- 6.29 x 10-8 NH4+ 5.68 x 10 -10 HCl >> 1 ​ Calculate...

Acid Ka

H2PO4- 6.29 x 10-8

NH4+ 5.68 x 10 -10

HCl >> 1 ​

Calculate the pH of a 0.150 M solution of Na2HPO4.

Calculate the pH of a 0.150 M solution of NH4NO3.

Calculate the pH of a 0.150 M solution of KCl.

Homework Answers

Answer #1

1) Na2HPO4 is salt of weak acid H2PO4- and strong base NaOH

for such salts

pH = 1/2 [pKw + pKa + logC]

pKa = - log Ka = - log [6.29 x 10-8] = 7.20

pH = 1/2 [14 + 7.20 + log 0.150]

pH = 10.19

2) NH4NO3 is salt of weak base and strong acid

pKa of NH4+ = - log [5.68 x 10-10] = 9.24

pKb = 14 - 9.24 = 4.76

pOH = 1/2 [pKw + pKb + log C]

pOH = 1/2 [14 + 4.76 + log 0.150]

pOH = 8.97

pH = 14 - 8.97

pH = 5.03

3) KCl is salt of strong acid and strong base

such salts pH always 7.0

pH = 7.0

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
SALTS: For each of the following salt solutions, record the experimental pH, determine the theoretical pH...
SALTS: For each of the following salt solutions, record the experimental pH, determine the theoretical pH (show your work) and then determine the percent error between the two (show your work). For errors of greater than 20%, provide an explanation for the deviation. 1) 0.15 M KCl Solution; experimental pH = 6.82 Acid Ka HCL >>1 Calculated pH: _____________ Actual pH: ______________ Percent error: ____________ 2) 0.15 M Na2HPO4 Solution; experimental pH = 8.90 Acid Ka H3PO4 7.1 x 10-3...
Calculate the pH of a 0.150 M solution of benzoic acid (C6H5CO2H) if   Ka= 6.3 x...
Calculate the pH of a 0.150 M solution of benzoic acid (C6H5CO2H) if   Ka= 6.3 x 10-5for the acid.    C6H5CO2H(aq)     ⇌    H+(aq) + C6C5CO2-(aq)
Given the following Ka values: H3PO4 <--> H2PO4- + H+, Ka = 7.11*10-3 H2PO4- <--> HPO42-...
Given the following Ka values: H3PO4 <--> H2PO4- + H+, Ka = 7.11*10-3 H2PO4- <--> HPO42- + H+, Ka = 6.34*10-8 HPO42- <--> PO43- + H+, Ka = 4.22*10-13, a) please calculate the pH of the solution at the first equivalence point for the above titration. b) In the same titration as described above, what's the pH of the solution after 15.00 mL of NaOH was added?
Phosphoric acid is a triprotic acid, and the Ka values are given below. Calculate pH, pOH,...
Phosphoric acid is a triprotic acid, and the Ka values are given below. Calculate pH, pOH, [H3PO4], [H2PO4 2-], [HPO4 -], and [PO4 3-] at equilibrium for a 5.00 M phosphoric acid solution. Ka1 = 7.5 x 10^-3 Ka2 = 6.2 x 10^-8 Ka3 = 4.2 x 10^-13
1.) If the Ka of a monoprotic weak acid is 4.5 x 10^-6, what is the...
1.) If the Ka of a monoprotic weak acid is 4.5 x 10^-6, what is the pH of a 0.30M solution of this acid? 2.) The Ka of a monoprotic weak acid is 7.93 x 10^-3. What is the percent ionization of a 0.170 M solution of this acid? 3.) Enolugh of a monoprotic acid is dissolved in water to produce a 0.0141 M solution. The pH of the resulting solution is 2.50. Calculate the Ka for the acid.
Caclulate the pH of: a 0.200 M solution of NaCH3COO a 0.200 M solution of NH4NO3...
Caclulate the pH of: a 0.200 M solution of NaCH3COO a 0.200 M solution of NH4NO3 a 0.200 M solution of NaCl CH3COOH 1.76 * 10-3 NH4+ 5.68 * 10-10 HCl >1 Predict whether the following would be acidic, basic, or neutral in solution: 0.5 M NaCN 0.5 M KCl 0.5 M Ba(NO3)2 0.5 M NH4C2H3O2
The Ka of propanoic acid (C2H5COOH) is 1.34 x 10-5. Calculate the pH of the solution...
The Ka of propanoic acid (C2H5COOH) is 1.34 x 10-5. Calculate the pH of the solution and the concentrations of C2H5COOH and C2H5COO- in a 0.297 M propanoic acid solution at equilibrium. pH = _____ [C2H5COOH] = _____ M [C2H5COO-] = ______ M
The Ka for hypochlorous acid, HClO, is 3.0 x 10-8.Calculate the pH after 13.0 mL of...
The Ka for hypochlorous acid, HClO, is 3.0 x 10-8.Calculate the pH after 13.0 mL of 0.100 M NaOH have been added to 30.0 mL of 0.100 M HClO
Calculate the pH of a 0.39 M CH3COOLi solution. Ka for acetic acid= 1.8 x 10^-5
Calculate the pH of a 0.39 M CH3COOLi solution. Ka for acetic acid= 1.8 x 10^-5
1. Valeric acid is a monoprotic acid with a Ka value of 1.44 x 10-5. A...
1. Valeric acid is a monoprotic acid with a Ka value of 1.44 x 10-5. A student prepared 100.0 mL of a 0.200 M solution of valeric acid. What is the pH of the valeric acid solution?