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scoray [572]
3 years ago
9

Hii pls helpnme to write out the ionic equation ​

Chemistry
1 answer:
Oxana [17]3 years ago
3 0

Answer:

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

Explanation:

According to this question, sodium carbonate reacts with sulfuric acid to form aqueous sodium sulfate, carbon dioxide and water. The balanced chemical equation is as follows:

Na2CO3(aq) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l)

- Next, split compounds that are aqueous into ions.

2Na+(aq) + CO32-(aq) + 2H+(aq) + SO42-(aq) → 2Na+(aq) + SO42-(aq) + CO2(g) + H2O(l)

- Next, we cancel out the spectator ions, which are ions that remain the same in the reactants and products side of a chemical reaction. The spectator ions in this equation are 2Na+(aq) and SO42-(aq).

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

- Hence, the balanced ionic equation is as follows:

CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)

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A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
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NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

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pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

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