The titration reaction can be represented as mgcl2(aq) 2 naoh(aq) → mg(oh)2(s) 2 nacl(aq) the solubility equilibrium is written as mg(oh)2(s) ⇌ mg2 (aq) 2 oh-(aq) kemmi is trying to teach noke how to calculate the concentration of the mg2 cation at the titration endpoint. she explains that the initial mgcl2 concentration must be multiplied by the volume of mgcl2 pipetted and divided by the total volume at the endpoint. (note: the calculation page in the lab manual has a similar equation for the lead cation.) noke do her calculation. for the first titration in the set the initial concentration of mgcl2(aq) was 0.0050 m, the volume measured was 100.00 ml, and the total volume of solution at the endpoint was 104.12 ml. calculate the concentration of mg2 at the endpoint. (answer with 2 significant digits and units of m)

The concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M

Explanation:

The titration reaction involved: MgCl₂(aq) + 2 NaOH(aq) → Mg(OH)₂(s) + 2 NaCl(aq)

and the solubility equilibrium for the product: Mg(OH)₂(s) ⇌ Mg²⁺ (aq) + 2 OH⁻(aq)

Given: Concentration of MgCl₂: M₁ = 0.005 M, volume of MgCl₂: V₁ = 100 mL, Total volume of solution at end point = 104.12 mL

Concentration of Mg²⁺ cation at end point: M₂ = ?M

Now, to calculate the concentration of Mg²⁺ cation at end point, we use the equation:    M₁ × V₁ = M₂ × V₂

⇒ M₂ = M₁ × V₁ ÷ V₂

⇒ M₂ = (0.005 M × 100 mL) ÷ 104.12 mL

⇒ M₂ = 0.0048 M (2 significant digits)

Therefore, the concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.