what is the ph of a solution in which 250.0 ml of .200 M HNO3 is added to 280.0 ml of .150 Ba(OH )
To find the pH of the solution, we need to determine the concentration of the resulting solution after the reaction between HNO3 and Ba(OH)2.
1. First, let's determine the moles of HNO3 and Ba(OH)2 used:
Moles of HNO3 = volume (L) × concentration (mol/L)
= 250.0 mL × 0.200 mol/L
= 0.0500 mol
Moles of Ba(OH)2 = volume (L) × concentration (mol/L)
= 280.0 mL × 0.150 mol/L
= 0.0420 mol
2. HNO3 and Ba(OH)2 react in a 1:2 ratio to form one mole of water (H2O) and one mole of barium nitrate (Ba(NO3)2):
HNO3(aq) + 2 Ba(OH)2(aq) → Ba(NO3)2(aq) + 2 H2O(l)
Since HNO3 is the limiting reagent (0.0500 mol), it will be fully consumed, yielding 0.0500 mol of Ba(NO3)2.
3. Next, let's calculate the concentration of Ba(NO3)2 in the final volume of the solution:
Total volume of the solution = volume of HNO3 + volume of Ba(OH)2
= 250.0 mL + 280.0 mL
= 530.0 mL
= 0.530 L
Concentration of Ba(NO3)2 = moles of Ba(NO3)2 / total volume of solution
= 0.0500 mol / 0.530 L
≈ 0.0943 M
4. Now that we have the concentration of Ba(NO3)2, we can determine its pOH. The pOH is the negative logarithm (base 10) of the hydroxide ion concentration (OH-) in the solution.
pOH = -log[OH-]
In this case, Ba(NO3)2 is a strong electrolyte and fully dissociates into its ions, so the concentration of OH- (hydroxide ions) is twice the concentration of Ba(NO3)2: [OH-] = 2 × 0.0943 M = 0.1886 M
pOH = -log(0.1886) ≈ 0.725
5. Finally, we can calculate the pH of the solution using the relation: pH + pOH = 14.
pH = 14 - pOH
= 14 - 0.725
≈ 13.275
Therefore, the pH of the solution is approximately 13.275.