Calculate the concentrations of all species in a 0.100 M H3PO4 solution.
Expert Solution & Answer
Interpretation Introduction
Interpretation:
The concentration of all species in a 0.100 M H3PO4 solution has to be calculated
Concept Information:
Acid ionization constantKa:
The equilibrium expression for the reaction HA(aq)⇄H+(aq)+A-(aq) is given below.
Ka=[H+][A-][HA]
Where Ka is acid ionization constant, [H+] is concentration of hydrogen ion, [A-] is concentration of acid anion, [HA] is concentration of the acid
Diprotic and polyprotic acids:
Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids. These acids lose one proton at a time by undergoing successive ionizations.
For diprotic acids, the successive ionization constants are designated as Ka1andKa2
For triprotic acids, the successive ionization constants are designated as Ka1,Ka2andKa3
Answer to Problem 15.122QP
The concentration of all species in a 0.100 M H3PO4 solution
[H+]= [H2PO4-]=0.0239 M [H3PO4]=0.076 M[HPO42-]=0.076 M [HPO42-]=6.2×10-8M[PO43-]=1.2×10-18M
Explanation of Solution
Given data:
The concentration of H3PO4 solution is 0.100 M
Species ofH3PO4
H3PO4 is a triprotic acid.
The species that are present in H3PO4 are H3PO4, H+, H2PO4-, HPO42- and PO43-.
The concentration of these species will be represented as [H3PO4], [H+], [H2PO4-],[HPO42-] and [PO43-] and can be calculated as follows.
First ionization ofH3PO4
H3PO4(aq)→ H+(aq) +H2PO4-(aq)
Initial (M)
0.100
-x
(0.100-x)
0.00
0.00
Change (M)
+x
+x
Equilibrium (M)
x
x
The Ka1 for phosphoric acid is 7.5×10−3
Ka1=[H+][H2PO4-]H3PO47.5×10−3=(x)20.100−x
In this case we probably cannot say that 0.100−x≈~0.100 due to the magnitude of Ka
We obtain the quadratic equation:
x2+(7.5×10−3)x−(7.5×10−4)=0
The positive root is x = 0.00239 M. We have:
[H+]= [H2PO4-]=0.0239 M [H3PO4]=(0.100-0.0239) M=0.076 M
Using the Nernst equation to calculate nonstandard cell voltage
A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction:
MnO2 (s)+4H* (aq)+2Cr²+ (aq) → Mn²+ (aq)+2H₂O (1)+2Cr³+ (aq)
+
2+
2+
3+
Suppose the cell is prepared with 7.44 M H* and 0.485 M Cr²+ in one half-cell and 7.92 M Mn² and 3.73 M Cr³+ in the other.
Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
☐
x10
μ
Х
5
?
000
日。
Calculating standard reaction free energy from standard reduction...
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG° for the following redox reaction.
Be sure your answer has the correct number of significant digits.
NO (g) +H₂O (1) + Cu²+ (aq) → HNO₂ (aq) +H* (aq)+Cu* (aq)
kJ
-
☐ x10
x10
olo
18
Ar
Calculating the pH of a weak base titrated with a strong acid
b
An analytical chemist is titrating 116.9 mL of a 0.7700M solution of aniline (C6H5NH2) with a 0.5300M solution of HNO3. The pK of aniline is 9.37.
Calculate the pH of the base solution after the chemist has added 184.2 mL of the HNO 3 solution to it.
Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added.
Round your answer to 2 decimal places.
pH = ☐
☑
5
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