Calculate the molarity of a solution of HNO3 that contains 12.6 grams HNO3 in

Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]

ohaa [14]

Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.021 M.
Ka(HCN) = 4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.021 M - x..
2.04·10⁻⁵ = x² / (0.021 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.

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2 years ago

If you are given a 0. 31 g piece of sodium metal to react with water, how many moles of hcl would it take to neutralize the sodi

Sphinxa [80]

The moles of HCl to neutralize the sodium hydroxide produced is 0.0135 mole.

Neutralization or neutralization is a chemical response wherein an acid and a base react quantitatively with each other. In a reaction in water, neutralization outcomes in there being no excess of hydrogen or hydroxide ions gift in the answer.

calculation:-

2Na + 2H₂O -----> 2NaOH + H₂

2 mol or 46g of Na produces 80 grams of NaOH

∴ 0.31 g of Na will produce = 80/46 × 0.31

= 0.54 gram of NaOH.

mol of NaOH = 0.54/40

= 0.0135

Since both Hcl and NaOH have the same valance factor,

1 mole NaOH is needed to neutralize 1 mol HCl

∴ 0.0135 mole of NaOH will need = 0.0135 mole of HCl

mass = 0.0135 × 36.5

= 0.493 grams of HCL.