Reaction | KOH(s) | --> | K+(aq) | OH-(aq) |
Given | 10.0 g | |||
MW | 56.1056 g/mol | |||
mole | 0.1782 mol | |||
produce | 0.1782 mol | 0.1782 mol |
Reaction | HCl(g) | + | H2O | -- | H3O+(aq) | + | Cl-(aq) |
given | 10.00 g | excess | |||||
MW | 36.4606 g/mol | ||||||
mole | 0.2742 mol | ||||||
produce | 0.2742 | 0.2742 | |||||
concentration | 1.097 M | 1.097 M |
reaction | HCl(aq) | + | H2O | --> | H3O+ | + | Cl- |
given | 10.0 g | excess | |||||
MW | 36.46 g/mol | since in | |||||
mole | 0.2746 mol | water | |||||
produce | 0.2743 | 0.3743 mole |
NaCl(s)
--> Na+(aq) + Cl-(aq)
mole NaCl = 0.08555 moles
[Na+] = 0.8555 M
[Cl-] = 0.8555 M
No solid present
AgNO3(s)
--> Ag+(aq) + NO3-(aq)
mole AgNO3 = 0.02943 moles
[Ag+] = 0.2943 M
[NO3-] = 0.2943 M
No solid present
Reaction | Na+(aq) | + | Cl-(aq) | + | Ag+(aq) | + | NO3-(aq) | --> | AgCl(s) | + | Na+(aq) | + | NO3-(aq) |
Start (moles) | 0.08555 | 0.08555 | 0.02943 | 0.02943 | |||||||||
Use (moles) | spectator ion | 0.02943 | 0.02943 | spectator ion | |||||||||
Make (moles) | 0.02943 | ||||||||||||
Final (moles) | 0.08555 | 0.05612 | 0 | 0.02943 | 0.02943 | ||||||||
Final (molarity) | 0.4278 M | 0.2806 M | 0 M | 0.1472 M | solid | ||||||||
Mass (g) | 4.218 g |
The important point to solving this problem is that in a titration, there is no limiting reagent. So for this reaction where:
HCl(aq) + NaOH(aq) --> H2O + NaCl(aq)
Given 15.36 mL of 0.3215 M HCl, this solution contains 4.938x10-3 moles HCl.
Therefor the sodium hydroxide solution must contain 4.938x10-3 moles NaOH
Therefor [NaOH] = 0.1975 M