Ask Question
19 October, 02:18

What is the pH when 5.0 mL of 0.010 M barium hydroxide is added to 20.0 mL of 0.020 M nitric acid

12.08

1.92

12.15

1.85

+4
Answers (1)
  1. 19 October, 03:50
    0
    Answer is: pH is 1.92.

    Balanced chemical reaction: Ba (OH) ₂ + 2HNO₃ → Ba (NO₃) ₂ + 2H₂O.

    V (Ba (OH) ₂) = 5.0 mL : 1000 mL/L.

    V (Ba (OH) ₂) = 0.005 L; volume of barium hydroxide.

    c (Ba (OH) ₂) = 0.010 M; molarity of barium hydroxide.

    n (Ba (OH) ₂) = V (Ba (OH) ₂) · c (Ba (OH) ₂).

    n (Ba (OH) ₂) = 0.00005 mol; amount of barium hydroxide.

    n (HNO₃) = 0.0004 mol; amoun of nitric acid.

    From balanced chemical reaction: n (Ba (OH) ₂) : n (HNO₃) = 1 : 2.

    Δn (HNO₃) = 0.0004 mol - 0.0001 mol.

    Δn (HNO₃) = 0.0003 mol; excess of nitri acid.

    Δc (HNO₃) = 0.0003 mol : 0.025 L.

    Δc (HNO₃) = 0.012 M.

    pH = - log (0.012 M).

    pH = 1.92.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “What is the pH when 5.0 mL of 0.010 M barium hydroxide is added to 20.0 mL of 0.020 M nitric acid 12.08 1.92 12.15 1.85 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers