Write the equation for the dissolution of sodium carbonate in water as found in your laboratory guide.

1 answer:

3 0

Na₂CO₃(s) → 2Na⁺(aq) + CO₃²⁻(aq)
The sodium carbonate formed from a strong base and a weak acid. Hydrolysis is subjected to the anion of a weak acid.

CO₃²⁻ + H₂O ⇄ HCO₃⁻ + OH⁻
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻

pH>7 alkaline solution

2Na⁺ + CO₃²⁻ + 2H₂O ⇄ 2Na⁺ + 2OH⁻ + H₂CO₃

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A 100.0 mL solution containing 0.923 gof maleic acid (MW=116.072 g/mol) is titrated with 0.265 M KOH. Calculate the pH of the so

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Answer:

pH = 9,57

[M²⁻] = 7,948x10⁻²M

[HM⁻] = 4x10⁻⁵M

[H₂M] = 0M

Explanation:

The moles of maleic acid presents in the solution are:

0,923g× $\frac{1mol}{116,072g}$ =7,952x10⁻³moles of H₂M

60,0mL of 0,265M KOH are:

0,0600L× $\frac{0,265mol}{1L}$ =0,0159 moles of KOH

The reactions of maleic acid (H₂M) and then with HM⁻ are:

H₂M + KOH → HM⁻ + H₂O + K⁺ (1)

HM⁻ + KOH → M²⁻ + H₂O + K⁺ (2)

For a complete transformation of H₂M in HM⁻ there are necessaries 7,952x10⁻³moles of KOH. As the moles of KOH are 0,0159 moles, the restant moles are:

0,0159 - 7,952x10⁻³ = 7,948x10⁻³ moles of KOH

By (2), the moles produced of M²⁻ are the same as moles of KOH, 7,948x10⁻³ moles, and moles of HM⁻ are:

7,952x10⁻³ - 7,948x10⁻³ = 4x10⁻⁶ moles of HM⁻

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [M²⁻] /[HM⁻]

pH = 6,27 + log₁₀ 7,948x10⁻³ / 4x10⁻⁶

pH = 9,57

The moles of M²⁻ are 7,948x10⁻³ and volume of the solution is 0,1000L,

[M²⁻] = 7,948x10⁻²M

Moles of HM⁻ are 4x10⁻⁶:

[HM⁻] = 4x10⁻⁵M

And there is not H₂M: