Given:
Diprotic weak acid H2A:
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m
Balanced chemical equation:
H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x
ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)
solve for x and determine the concentration at equilibrium.
Answer:
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Answer:
130ml of HCl(36%) in 4.90L solution => pH = 1.50
Explanation:
Need 4.90L of HCl(aq) solution with pH = 1.5.
Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺
[HCl(36%)] ≅ 12M in HCl
(M·V)concentrate = (M·V)diluted
12M·V(conc) = 0.032M·4.91L
=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.
Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.
Liquidity metric is the other name
