POH = 14 - pH
pOH = -log [OH-]
Answer:
<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>
Explanation:
No. of moles =
= = 0.001245 moles
Concentration of KHP (C1) in litres = n/v
= = 0.062 mol/L
We know that:
=
where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.
Since mole ratio is 1 : 1.
1 mole of NaOH - 40g
0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g
⇒0.0498g of NaOH was used during the titration
<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>
<u><em>= 2.49g/L.</em></u>
You may find bellow the balanced chemical equations.
Explanation:
Molecular equations:
3 Sr(NO₃)₂ (aq) + 2 K₃PO₄ (aq) → Sr₃(PO₄)₂ (s) + 6 KNO₃ (aq)
2 NaOH (aq) + Ni(NO₃)₂ (aq) = Ni(OH)₂ (s) + 2 NaNO₃ (aq)
Ionic equations:
3 Sr²⁺ (aq) + 6 NO₃⁻ (aq) + 6 K⁺ (aq) + 2 PO₄³⁻ (aq) → Sr₃(PO₄)₂ (s) + 6 K⁺ (aq) + 6 NO₃⁻ (aq)
2 Na⁺ (aq) + 2 OH⁻ (aq) + Ni²⁺ (aq) + 2 NO₃⁻ (aq) = Ni(OH)₂ (s) + 2 Na⁺ (aq) + 2 NO₃⁻ (aq)
To get the net ionic equation we remove the spectator ions:
3 Sr²⁺ (aq) + 2 PO₄³⁻ (aq) → Sr₃(PO₄)₂ (s)
2 OH⁻ (aq) + Ni²⁺ (aq) = Ni(OH)₂ (s)
Learn more about:
net ionic equations
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