Answer:
Molarity NaOH = 0.85M (2 sig figs)
Explanation:
48.0ml(0.220M H₂SO₄) + 25ml(Xmolar NaOH)H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
2(molarity x volume) H₂SO₄ = (molarity x volume) NaOH
2(0.220M x 48.0ml) = 25.0ml x Molarity NaOH
Molarity NaOH = 2(0.220M x 48.0ml)/25.0ml = 0.8448M ≅ 0.85M (2 sig figs)
<span>M(NO3)2 ==> [M2+] + 2 [NO3-]
0.202 M ==> 0.202 M
M(OH)2 ==> [M2+] + 2[OH-]
5.05*10^-18 ===> s + [2s]^2
5.05*10^-18 ===> 0.202 + [2s]^2
5.05*10^-18 = 0.202 * 4s^2
4s^2 = 25*10^-18
s^2 = 6.25*10^-18
s = 2.5*10^-9
So, the solubility is 2.5*10^-9</span>
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Notice q=3/2, is half of the original q = 3(<span>1/2</span>)<span>t/28.8
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