Complete this table of values for four aqueous solutions at 25 degrees C.

Question

Question:

Two uniform spheres, each with mass MM and radius RR, touch one another.What is the magnitude of their gravitational force of attraction?. Express your answer in terms of the variables MM and RR and appropriate constants.

Solution A: H^+ = 7.3 * 10^-13, OH^- = ?, pH = ?, pOH = ?

Solution B: H^+ = ?, OH^- = 0.078, pH = ?, pOH = ?

Solution C: H^+ = ?, OH^- = ?, pH = 5.89, pOH = ?

Solution D: H^+ = ?, OH^- = ?, pH = ?, pOH = 11.99

Answer:

Figuring for pH, pOH, [H+]and [OH−]

The autoionization for water is given by
[H+] [OH−]=Kw
where Kw=1.0×10−14
The pH-pOH connection is given by
pH + pOH=14
The connection among pH and hydrogen particle focus is given by
[H+] =10−pH
also,
pH= − log[H+]
While the connection among pOH and the hydroxide particle focus is given by
[OH−]=10−pOH
also,
pOH= − log[OH−]
Answer and Explanation:
Arrangement A
[H+] = 7.3×10−13
Tackling for [OH−]
[OH−] = 1.0×10−14/[H+]
[OH−] = 1.0×10−14/7.3×10−13
[OH−] = 1.4×10−2
Tackling for pH
pH=−log[H+]
pH=−log7.3×10−13
pH=12.1
Tackling for pOH
pH + pOH =14
pOH=14−pH
pOH=14−12.1
pOH=1.9
Arrangement B
[OH−] = 0.078
Tackling for [H+]
[H+] =1.0×10−14/[OH−]
[H+] = 1.0×10−14/0.078
[H+] = 1.28×10−13
Tackling for pH
pH=−log[H+]
pH=−log1.28×10−13
pH=12.89
Tackling for pOH
pH+ pOH =14
pOH =14−pH
pOH =14−12.89
pOH=1.11
Arrangement C
pH=5.89
Tackling for pOH
pH+ pOH = 14
pOH = 14−pH
pOH =14−5.89
pOH = 8.11
Tackling for [H+]
[H+] = 10−pH
[H+] = 10−5.89
[H+] = 1.29×10−6
Tackling for [OH−]
[OH−]=1.0×10−14/[H+]
[OH−]=1.0×10−14/1.29×10−6
[OH−]=7.75×10−9
Arrangement D
pOH=11.99
Tackling for pH
pH+pOH=14
pH=14−pOH
pH=14−11.99
pH=2.01
Tackling for [H+]
[H+]=10−pH
[H+]=10−2.01
[H+]=9.77×10−3
Tackling for [OH−]
[OH−]=1.0×10−14/[H+]
[OH−]=1.0×10−14/9.77×10−3
[OH−]=1.02×10−

 

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