Question

13. Copper(II) chloride and silver acetate Acid/Base:

Answer 1

Answer:

13. $CuCl_2 \hspace{0.1cm}+ CH_3COOAg -> AgCl \hspace{0.1cm} + (CH_3COO)_2Cu$

This is a double displacement reaction, not an acid-base reaction

14.$HF\hspace{0.1cm}+Ba(OH)_2 ->\hspace{0.1cm}H_2O\hspace{0.1cm}+BaF_2$

$Ba(OH)_2 -> Base$

$HF -> Acid$

15. $HCl\hspace{0.1cm}+Mg(OH)_2->\hspace{0.1cm}H_2O\hspace{0.1cm}+MgCl_2$

$MgCl_2 -> Base$

$HCl -> Acid$

16.$CH_3COOH\hspace{0.1cm}+LiOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+CH_3COOLi$

$LiOH -> Base$

$CH_3COOH -> Acid$

17.$H_2SO_4\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaHSO_4$

$NaOH -> Base$

$H_2SO_4-> Acid$

18. $H_2SO_4\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaHSO_4$

$NaOH -> Base$

$H_2SO_4-> Acid$

19.$H_2SO_4\hspace{0.1cm}+NaHCO3->\hspace{0.1cm}H_2O\hspace{0.1cm}+CO_2+\hspace{0.1cm}Na_2SO_4$

$NaHCO_3-> Base$

$H_2SO_4-> Acid$

20.$H_3PO_4\hspace{0.1cm}+Zn(OH)_4->\hspace{0.1cm}H_2O\hspace{0.1cm}+Zn_3(PO_4)_4$

$Zn(OH)_4> Base$

$H_3PO_4-> Acid$

21. $HF\hspace{0.1cm}+Na_2SO_3->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaF\hspace{0.1cm}+\hspace{0.1cm}SO_2$

$NaSO_2> Base$

$HF-> Acid$

22.$HCl\hspace{0.1cm}+Na_2SO_3->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaCl\hspace{0.1cm}+\hspace{0.1cm}SO_2$

$NaSO_2> Base$

$HCl-> Acid$

23.$HCl\hspace{0.1cm}+Na_2S->\hspace{0.1cm}H_2S\hspace{0.1cm}+NaCl$

$Na_2S> Base$

$HCl-> Acid$

24. $NH_4Cl\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaCl+{0.1cm}NH_3$

$NaOH> Base$

$NH_4Cl-> Acid$

Explanation:

All acid-base reaction have the same products, $H_2O$ and a salt. Whit this in mind the acids would the compounds that produces the hydronium ion $(H^+)$ and the bases would be the compounds that produces the hydroxide ion $(OH^-)$