Question #130a3

Question

Question #130a3

1 Answer

Impact of this question

1432 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-02T16:40:29

$N a H S O_{4}$ $NaHSO_4$ produces $H^{+}$ $H^+$ ion in its aqueous solution, which combines with $C O_{3}^{2} -$ $CO_3^2-$ ions (formed on ionisation of $N a_{2} C O_{3}$ $Na_2CO_3$ ) to form $H_{2} C O_{3}$ $H_2CO_3$ ,an unstable weak acid.This acid readily decomposes to form $C O_{2}$ $CO_2$ gas.

$2 N a H S O_{4} \to 2 N a^{+} + 2 H^{+} + 2 S O_{4}^{2-}$ $2NaHSO_4->2Na^+ + 2H^+ + 2SO_4^"2-"$

$N a_{2} C O_{3} \to 2 N a^{+} + C O_{3}^{2-}$ $Na_2CO_3->2Na^+ + CO_3^"2-"$

$C O_{3}^{2-} + 2 H^{+} \to H_{2} C O_{3}$ $CO_3^"2-" + 2H^+ -> H_2CO_3$

$H_{2} C O_{3} \to H_{2} O + C O_{2}$ $H_2CO_3->H_2O + CO_2$

Equation of the net reaction

$N a_{2} C O_{3} + 2 N a H S O_{4} \to 2 N a_{2} S O_{4} + H_{2} O + C O_{2}$ $Na_2CO_3+2NaHSO_4->2Na_2SO_4+H_2O+CO_2$

Science

Math

Humanities

... and beyond

Question #130a3

1 Answer

Related questions

Impact of this question