A sample of pure tin metal is dissolved in nitric acid to produce 20.00 mL of solution containing Sn2+. When this tin solution is titrated, a total of 39.8 mL of 0.140 mol/L KMnO4is required to reach the equivalence point.
b.What is the concentration of the Sn2+ solution?
5 years ago
Answered By Elijah A
I like to begin every chemistry question by writing out the reactions and everything I know about them. The question says that a metal is dissolved in an acid, so we know that our products are going to be a salt and hydrogen. We use this purely to get the tin solution chemical compound.
Sn(s)+ 2HNO3 (aq) --> H2 (g) + Sn(NO3)2 (aq)
We also have a titration with Sn(NO3)2 (aq) and KMnO4 (aq) which turns into a redox question. So starting from the components to identify your strongest reducing agent and strongest oxidizing agent:
Sn2+ NO3- K+ MnO4- H+
and using the Chemistry 30 data booklet we get the following two reactions for SOA and SRA, respectively:
MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O
Sn4+ + 2e- --> Sn2+
Obviously flipping the reducing reaction and combining to get rid of the electrons on both sides (multiply SOA reaction by 2 and SRA reaction by 5) we get:
2MnO4- + 16H+ + 5Sn2+ --> 2Mn2+ + 8H2O + 5Sn4+
Now it is simply stoich, we have 2 mols of MnO4- per 5 mols of Sn2+, so when we have n mols of MnO4- we will get n*(5/2) mols of Sn2+. In this case, $n=\frac{0.140mols}{1L}\times39.8mL\times\frac{1L}{1000mL}=\text{0.00557mols}$n=0.140mols1L×39.8mL×1L1000mL=0.00557mols and thus the number of mols of Sn2+$=\text{0.00557mols}\times\frac{5}{2}=\text{0.0139mols}$=0.00557mols×52=0.0139mols .
The question asks for concentration so we have to divide our mols by volume (20.00mL given in question):
$C=\text{0.0139mols}\times\frac{1}{20.00mL}\times\frac{1000mL}{1L}=\text{0.6965}$C=0.0139mols×120.00mL×1000mL1L=0.6965 mol/L. That is our final answer. I hope this helped and let me know if I can answer anymore questions you might have. Apologies for my formatting, I am still getting used to the site.
5 years ago
Answered By Elijah A
I like to begin every chemistry question by writing out the reactions and everything I know about them. The question says that a metal is dissolved in an acid, so we know that our products are going to be a salt and hydrogen. We use this purely to get the tin solution chemical compound.
Sn(s)+ 2HNO3 (aq) --> H2 (g) + Sn(NO3)2 (aq)
We also have a titration with Sn(NO3)2 (aq) and KMnO4 (aq) which turns into a redox question. So starting from the components to identify your strongest reducing agent and strongest oxidizing agent:
Sn2+ NO3- K+ MnO4- H+
and using the Chemistry 30 data booklet we get the following two reactions for SOA and SRA, respectively:
MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O
Sn4+ + 2e- --> Sn2+
Obviously flipping the reducing reaction and combining to get rid of the electrons on both sides (multiply SOA reaction by 2 and SRA reaction by 5) we get:
2MnO4- + 16H+ + 5Sn2+ --> 2Mn2+ + 8H2O + 5Sn4+
Now it is simply stoich, we have 2 mols of MnO4- per 5 mols of Sn2+, so when we have n mols of MnO4- we will get n*(5/2) mols of Sn2+. In this case, $n=\frac{0.140mols}{1L}\times39.8mL\times\frac{1L}{1000mL}=\text{0.00557mols}$n=0.140mols1L ×39.8mL×1L1000mL =0.00557mols and thus the number of mols of Sn2+ $=\text{0.00557mols}\times\frac{5}{2}=\text{0.0139mols}$=0.00557mols×52 =0.0139mols .
The question asks for concentration so we have to divide our mols by volume (20.00mL given in question):
$C=\text{0.0139mols}\times\frac{1}{20.00mL}\times\frac{1000mL}{1L}=\text{0.6965}$C=0.0139mols×120.00mL ×1000mL1L =0.6965 mol/L. That is our final answer. I hope this helped and let me know if I can answer anymore questions you might have. Apologies for my formatting, I am still getting used to the site.