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The SI unit for acceleration is metres per second squared. - m/s^2.
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
Answer:
The answer is given below.
Explanation:
We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.
AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M
HA → H+ + A-
Initial concentration: 0.1 → 3.98 ^10-3 + 0
equilibrium concentration: 0.1(1-α) → 3.98 * 10-3 + 0.1α 0.1α
pKa of chloroacetic acid is 2.9
-log(Ka) = 2.9
Ka = 1.26 * 10-3
From the equation, Ka = [H+] * [A-] / [HA]
1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)
Since α<<1, we assume 1-α = 1
Solving the equation, we have: α = 0.094
Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%
answer: 3g. 17kg+3 ÷ 0.25