When HCl is added to a saturated solution of PbCl2, the solubility of PbCl2 would decrease so precipitation would occur. The decrease in the solubility is due to the common ion effect or the presence of Cl- ions in both compounds.
Answer:
28.11g
Explanation:
Step 1:
Background understanding:
From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 atoms. This also indicates that 1 mole of helium (He) contains 6.02x10^23 atoms.
1 mole of He = 4g
Step 2:
Determination of the mass of He that contain 4.23x10^24 atoms. This is illustrated below:
4g of He contains 6.02x10^23 atoms.
Therefore Xg of He will contain 4.23x10^24 atoms i.e
Xg of He = (4x4.23x10^24)/6.02x10^23
Xg of He = 28.11g
Therefore, 28.11g of He contains 4.23x10^24 atoms
Answer:
pH = 5.35
Explanation:
Given 1.60 grams sodium acetate (NaOAc(aq))*** added to 50ml of 0.10M acetic acid (HOAc(aq)) solution.
Applying common ion effect keeping in mind that the addition of NaOAc provides the common-ion (OAc⁻).
HOAc(aq) ⇄ H⁺(aq) + OAc⁻(aq)
I 0.10m 1.32 x 10⁻³M ≈ ∅M* (1.6g/82.03g/mol) / 0.050L = 0.39M
C -x +x 0.39M + x ≈ 0.39M**
E 0.10M - x x 0.39M
≈ 0.10M
Ka = [H⁺][OAC⁻]/[HOAC] => [H⁺] = Ka·[HOAc] / [OAc⁻]
[H⁺] = (1.75 X 10⁻⁵)(0.10) / (0.39) = 4.5 x 10⁻⁶M
∴ pH = -log[H⁺] = -log(4.5 x 10⁻⁶) = -(-5.35) = 5.35
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* [H⁺] before adding NaOAc = SqrRt(Ka · [HOAc]) = SqrRt(1.75 x 10⁻⁵· 0.10) = 1.32 x 10⁻³M. Since this concentration value is so small, the initial [H⁺] is assumed to be zero molar (∅M).
** The added [H⁺] is negligible and dropped in the ICE table. That is, adding ~[H⁺] in the order of 10⁻³M does not change the H⁺ ion concentration sufficiently to affect problem outcome and is therefore dropped in the ICE table.
*** Acetic Acid and Sodium Acetate are frequently written HOAc and NaOAc where the OAc⁻ anion is the acetate ion (CH₃COO⁻) for brevity.
