Question

1-Each of the following metals reacts with nitric acid to form the metal chloride and hydrogen gas. Complete and bal- ance the reactions to show the formation of these products.
a. Zn (s) +HNO, (aq)->
b. Pb (s) +HNO3 (aq) -
c. Na (s) HNO3 (aq)- d. Al (s) HNO, (aq)-

2- Complete and balance these acid-base neutralization reactions: a. CSOH (aq) HBr (aq)
b. HBr (aq) Fe(OH)2 (s)
c. H,PO4 (aq) +KOH (aq)-

3-In a titration experiment, two drops of phenolphthalein are added to 25.0 mL of aq. NaOH solution, causing it to turn bright pink. A solution of 3.05-M hydrochloric acid is added to the solution. It is found that 5.2 mL of the acid solution is required to turn the solution colorless. What is the concentration of the original base solution?

Answer 1

Please note a) that there formation of nitrates (instead of chloride, as mentioned in question) .

b) Most of the metals when react with nitric acid (dilute or concentrated) does not liberate hydrogen gas, because nitric acid is very strong oxidising agent and so it oxidises the H gas into H2O and other oxides of nitrogen are also formed. Mg is the metal that can give hydrogen gas only with very dilute nitric acid.

So lets answer these questions

1 a) Zn(s) + 4HNO3(aq) _-----> Zn(NO3)2(aq) + 2NO2(g) + 2H2O

b) 3Pb(s) + 8 HNO3(aq) _-----> 3 Pb(NO3)2(aq) + 2NO(g) + 4H2O

c) Na(s) + HNO₃(aq)   _-----> NaNO3(aq) + NO(g) + N₂O(g) + H₂

d) Al(s) + 6HNO₃ (aq) _-----> Al(NO₃)₃(aq) + 3NO2(g) + 3H₂O

2 a) CsOH(aq) + HBr(aq) -----> CsBr(aq) + H₂O

b) Fe(OH)₂(aq) + 2HBr(aq) -----> FeBr₂(aq) + 2H₂O

c) H₃PO₄(aq) + 3KOH(aq) -------> K₃PO₄(aq) + 3H₂O

3) As the neautrallization reaction always occur by simple rule, one equivalent base reacts with one equivalent of acid, and in this case as the solution turns pink immediately, which indicates it to be basic, when it is neutralized by the acid then it turns colorless

for mono basic acids like HCl and monoacid bases like NaOH 1 equivalent = 1 mole

so, applying equation M_bV_b(base) = MaV_a(acid)

  M_bX25.0 mL = 3.05X5.2 mL

  M_b = 0.6344 M = 0.63 M (approx)

so, the initial concentration of the base was 0.63 M