Calculate the pH of 0.00010 M KOH. a. 5.0 b. 4.0 c. 3.0 d. 2.0 e. none of these

Question

Calculate the pH of 0.00010 M KOH. a. 5.0 b. 4.0 c. 3.0 d. 2.0 e. none of these Calculate the pH M Ba(OH)_2. a. 1.00 b. 12.7 c. 13.0 d. 14.0 e. none of these A 1.0-liter solution contains 1.0 mole of NaOH mixed with 1.0 mole of HClO_4. Determine its pH. a. 6.0 b. 6.5 c. 7.0 d. 7.5 e. none of these A 1.0 liter solution contains 1.00 mole of HNO_3 mixed with 1.10 moles of NaOH. Determine its pH. a. 1.0 b. 7.0 c. 13 d. 14 e. none of these A 1.0-liter solution contains 1.0 mole of HNO_3 mixed with 1.0 mole of NH_3. Determine its pH. a. 1.0 b. 4.6 c. 7.0 d. 9.4 e. none of these The pH of a 0.10 M aqueous solution is 7.0. The solute could be? a. NaOH b. NaOAc c. NH_3 d. NH_4 NO_3 e. none of these The pH of a 0.10 M solution is 8.9. The solute could be? a. NaNO_3 b. NaOAc c. NH_3 d. NH_4NO_3 e. none of these

Explanation / Answer

Solution:-

(17) KOH is a strong base and dissociate completely into its ions.

KOH(aq) ----> K+(aq) + OH-(aq)

Given concentration of KOH is 0.00010 M. So, OH- will also be 0.00010 M.

pOH = - log[OH-]

pOH = - log 0.00010 = 4

pH = 14 - 4

pH = 10

So, the correct choice is e. none of these.

(18) Ba(OH)2 is also a strong base and dissociates as....

Ba(OH)2(aq) -----> Ba2+(aq) + 2OH-(aq)

one mol of Ba(OH)2 gives 2 moles of OH-. So, the concentration of OH- will be two times the concentration of Ba(OH)2.

[OH-] = 2 x 0.0500 M = 0.100 M

pOH = - log 0.100

pOH = 1.0

pH = 14 - 1

pH = 13.0

So, the correct choice is c.

(19) we have 1.0 mole of NaOH and 1.0 mole of HClO4. Volume of solution is 1.0 L. So, the concentrations will also be 1.0 M for each.

NaOH is a strong base and HClO4 is a strong acid. Since, we have equal moles of these two so the solution will be neutral and hence the pH will be 7.

The, correct choice is c. 7.0

(20) we have 1.10 moles of NaOH and 1.00 moles of HNO3. The react in 1:1 mol ratio so the excess moles of NaOH = 1.10 - 1.00 = 0.10

Volume of solution = 1.0 L

molarity of excess NaOH = 0.10 mole/1.0 L = 0.10 M

pOH = - log 0.10 = 1.0

pH = 14 - 1.0 = 13.0

So, the correct choice is d.

(21) we have 1.0 mole of HNO3(strong acid) and 1.0 mole of NH3(weak base). So, the solution would be acidic.They react to form 1.0 mole of NH4+ that react with water as....

NH4+(aq) + H2O(l) <------> H3O+(aq) + NH3(aq)

I 1.0 0 0

C -X +X +X

E (1.0 - X) X X

Ka = [H3O+][NH3]/[NH4+]

Ka for NH4+ is 5.56 x 10-10

So, 5.56 x 10-10 = (X)2/(1.0 - X)

X on the bottom could be neglected as the value of Ka is very low. So,...

5.56 x 10-10 = (X)2/1.0

taking square root to both sides...

X = 2.36 x 10-5

pH = - log(H3O+] = - log(2.36 x 10-5)

pH = 4.6

So, the correct choice is b.

(22) pH of the solution is given as 7.0 that indicates the solution is neutral.

NaOH is a strong base and the pH for strong bases is very high(greater than 7).

NH3 is a weak base, so the pH will also be greater than 7 for this.

NaOAc (sodium acetate) is a salt of strong base (NaOH) and weak acid(CH3COOH), so the solution will be basic and the pH would be greater than 7.

NH4NO3 is a salt of weak base(NH3) and strong acid(HNO3). so, the solution will be acidic and the pH would be less than 7.

So, the only correct choice would be e. none of these.

(23) The pH of the solution is given as 8.9 that indicates the solution is basic.

NaNO3 is a salt of strong base and strong acid so the solution would be neutral and pH would be around 7.0.

NaOAc is a salt of strong base and weak acid so the pH would greater than 7. so, it would also be one of the choices. we will calculate the pH for this one also.

NH4NO3 is a salt of weak base(NH3) and strong acid(HNO3). so, the solution will be acidic and the pH would be less than 7.

NH3 is a weak base so it might be one of the correct choices. To validate this we could make the ice table and calculate the pH and see if it's same as the given one.

NH3(aq) + H2O(l) <-----> NH4+(aq) + OH-(aq)

I 0.10 0 0

C -X +X +X

E (0.10 - X) X X

Kb = [NH4+][OH-]/[NH3]

Kb for NH3 is 1.8 x 10-5

1.8 x 10-5 = (X)2/(0.10) (X is neglected on the bottom as the Kb value is very low)

on cross multiply...

X2 = 1.8 x 10-6

taking square root to both sides...

X = 1.3 x 10-3

pOH = - log 1.3 x 10-3 = 2.9

pH = 14 - 2.9 = 11.1

NH3 is not the right one since it gives different pH. Now let's make the ice table for NaOAc that is OAc-.

OAc-(aq) + H2O(l) <-----> HOAc(aq) + OH-(aq)

I 0.10 0 0

C -X +X +X

E (0.10 - X) X X

Kb = [HOAc] [OH-]/[OAc-]

Kb for 5.56 x 10-10

5.56 x 10-10 = (X)2/0.10 (X is also neglected here on the bottom as the dissociation constant value is low)

on cross multiply..

X2 = 5.56 x 10-11

taking square root to both sides..

X = 7.46 x 10-6

pOH = - log 7.46 x 10-6 = 5.1

pH = 14 - 5.1 = 8.9

So, the correct choice is b. NaOAc.

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