Calculate the pH and pOH of each of the following aqueous solutions of strong acd or base:

(a) 0.010 M HNO3(aq)
(b) ) 1.0 x 10-3M Ba(OH)2(aq)
(c) 10.0 mL of 0.22M KOH(aq) after dilution to 250 mL
(d) 100 mL aqueous solution containing 050 g of Ba(OH)2 (aq)

Here is how you do a and b.

a.
HNO3 is a strong acid; i.e., it ionizes 100%. Therefore, HNO3 ==> H^+ + NO3^- amd H^+ = 0.01
Then pH = -log(H^+) = -log(0.01) = -(-2) = 2

b.
Ba(OH)2 is a strong base and ionizes 100% as in Ba(OH)2 ==> Ba^2+ + 2OH^-
In a above you had only 1 H^+ in HNO3; here you have 2 OH^- in Ba(OH)2; therefore, (OH^-) is twice Ba(OH)2 so (OH^-) = 2*1E-3 = 2E-3
pOH = -log(OH^-) = -log(2E-3) = -(-2.699) = -2.699 which I would round to 2.7. To convert to pH use
pH + pOH = pKw = 14
pH + 2.7 = 14 and solve for pH.

c is done the same way but it's diluted; therefore (KOH) = 0.22M KOH x (10/250) = ? and go from there as in b above.

d. First determine M Ba(OH)2.
mols Ba(OH)2 = grams/molar mass, then
M = mols/L solution and proceed as in b.
Post your work if you get stuck.

first calcu the no. of mole of KOH

c=n/v
n=c.v =0.22*(10.0/1000) =0.0022 moles

Final conc.= 0.0022/(250/1000)= 0.0088M

pOH= -log[OH]= -log[0.0088M]=2.06
PH=14-2.06= 11.94

To calculate the pH and pOH of each of the given aqueous solutions, we need to use the concept of pOH and pH in relation to the concentration of H+ and OH- ions.

(a) 0.010 M HNO3(aq):
Since HNO3 is a strong acid, it will completely dissociate in water, yielding H+ ions and NO3- ions. As a result, the concentration of H+ ions is equal to the concentration of HNO3. Therefore, the concentration of H+ ions is 0.010 M.
To calculate the pH, we use the formula: pH = -log[H+]. In this case, pH = -log(0.010) ≈ 2.00.
Since pH + pOH = 14, we can calculate the pOH using pOH = 14 - pH. In this case, pOH = 14 - 2.00 ≈ 12.00.

(b) 1.0 x 10^-3 M Ba(OH)2(aq):
Ba(OH)2 is a strong base and it will completely dissociate in water, yielding Ba2+ ions and 2 OH- ions. Since the concentration of Ba(OH)2 is given as 1.0 x 10^-3 M, the concentration of OH- ions is 2 x 1.0 x 10^-3 = 2 x 10^-3 M.
To calculate the pOH, we use the formula: pOH = -log[OH-]. In this case, pOH = -log(2 x 10^-3) ≈ 2.70.
Since pH + pOH = 14, we can calculate the pH using pH = 14 - pOH. In this case, pH = 14 - 2.70 ≈ 11.30.

(c) 10.0 mL of 0.22 M KOH(aq) after dilution to 250 mL:
To calculate the pH and pOH after dilution, we need to consider the dilution factor. The dilution factor is given by the ratio of the final volume to the initial volume.
So, the dilution factor is (250 mL / 10 mL) = 25.
To calculate the new concentration, we divide the initial concentration by the dilution factor: 0.22 M / 25 ≈ 0.0088 M.
Since KOH is a strong base and it completely dissociates, the concentration of OH- ions is equal to the concentration of KOH, which is 0.0088 M.
To calculate the pOH, we use the formula: pOH = -log[OH-]. In this case, pOH = -log(0.0088) ≈ 2.06.
Since pH + pOH = 14, we can calculate the pH using pH = 14 - pOH. In this case, pH = 14 - 2.06 ≈ 11.94.

(d) 100 mL aqueous solution containing 0.50 g of Ba(OH)2(aq):
To calculate the concentration of Ba(OH)2, we need to convert the mass of Ba(OH)2 to moles and then divide it by the volume of the solution.
The molar mass of Ba(OH)2 is 171.34 g/mol.
The number of moles of Ba(OH)2 is 0.50 g / 171.34 g/mol ≈ 0.0029 mol.
The concentration of Ba(OH)2 is 0.0029 mol / 0.100 L = 0.029 M.
Since Ba(OH)2 is a strong base and it completely dissociates, the concentration of OH- ions is 2 x 0.029 = 0.058 M.
To calculate the pOH, we use the formula: pOH = -log[OH-]. In this case, pOH = -log(0.058) ≈ 1.24.
Since pH + pOH = 14, we can calculate the pH using pH = 14 - pOH. In this case, pH = 14 - 1.24 ≈ 12.76.

Therefore, the pH and pOH of each of the given solutions are:
(a) pH ≈ 2.00, pOH ≈ 12.00.
(b) pH ≈ 11.30, pOH ≈ 2.70.
(c) pH ≈ 11.94, pOH ≈ 2.06.
(d) pH ≈ 12.76, pOH ≈ 1.24.