pH and pOH of 0.00141 M HNO 3 is to be calculated. Concept Introduction: To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H + ] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution. Thus, pH = -log [H + ] And, at 25 ° C , the actual concentrations of the products are: [H + ][OH − ]=1 × 10 − 14 The mathematical product of both [H + ] and [OH − ] concentrations are always constant known as ion product constant ( K w ) [H + ][OH − ]=1 × 10 − 14 = K w The relation between pH and pOH can be derived from K w . pH+pOH=14
pH and pOH of 0.00141 M HNO 3 is to be calculated. Concept Introduction: To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H + ] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution. Thus, pH = -log [H + ] And, at 25 ° C , the actual concentrations of the products are: [H + ][OH − ]=1 × 10 − 14 The mathematical product of both [H + ] and [OH − ] concentrations are always constant known as ion product constant ( K w ) [H + ][OH − ]=1 × 10 − 14 = K w The relation between pH and pOH can be derived from K w . pH+pOH=14
To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H+] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution.
Thus, pH = -log [H+]
And, at 25°C , the actual concentrations of the products are:
[H+][OH−]=1×10−14
The mathematical product of both [H+] and [OH−] concentrations are always constant known as ion product constant ( Kw )
[H+][OH−]=1×10−14=Kw
The relation between pH and pOH can be derived from Kw .
pH+pOH=14
(a)
Expert Solution
Answer to Problem 20CR
pH of 0.00141 M HNO3 is 2.85 and pOH is 11.15.
Explanation of Solution
Concentration of H+ ion is 0.00141 M HNO3 in given solution. From this information, pH can be calculated.
Hence, pH and pOH of 0.00141 M HNO3 are 2.85 and 11.15 respectively.
(b)
Interpretation Introduction
Interpretation:
pH and pOH of 2.13×10−3M NaOH is to be calculated.
Concept Introduction:
To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H+] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution.
Thus, pH = -log [H+]
And, at 25°C , the actual concentrations of the products are:
[H+][OH−]=1×10−14
The mathematical product of both [H+] and [OH−] concentrations are always constant known as ion product constant ( Kw )
[H+][OH−]=1×10−14=Kw
The relation between pH and pOH can be derived from Kw .
pH+pOH=14
(b)
Expert Solution
Answer to Problem 20CR
pH of 2.13×10−3M NaOH is 11.33 and pOH is 2.67.
Explanation of Solution
Concentration of OH- ion is 2.13x10-3 M NaOH in given solution. From this information, pOH can be calculated.
Hence, pH and pOH 2.13×10−3M NaOH are 2.67 and 11.33 respectively.
(c)
Interpretation Introduction
Interpretation:
pH and pOH of 0.00515 M HCl is to be calculated.
Concept Introduction:
To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H+] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution.
Thus, pH = -log [H+]
And, at 25°C , the actual concentrations of the products are:
[H+][OH−]=1×10−14
The mathematical product of both [H+] and [OH−] concentrations are always constant known as ion product constant ( Kw )
[H+][OH−]=1×10−14=Kw
The relation between pH and pOH can be derived from Kw .
pH+pOH=14
(c)
Expert Solution
Answer to Problem 20CR
pH of 0.00515 M HCl is 2.29 and pOH is 11.71.
Explanation of Solution
Concentration of H+ ion is 0.00515 M HCl in given solution. From this information, pH can be calculated.
Hence, pH and pOH of 0.00515 M HCl are 2.29 and 11.71 respectively.
(d)
Interpretation Introduction
Interpretation:
pH and pOH of 5.65×10−5M Ca(OH)2 is to be calculated.
Concept Introduction:
To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H+] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution.
Thus, pH = -log [H+]
And, at 25°C , the actual concentrations of the products are:
[H+][OH−]=1×10−14
The mathematical product of both [H+] and [OH−] concentrations are always constant known as ion product constant ( Kw )
[H+][OH−]=1×10−14=Kw
The relation between pH and pOH can be derived from Kw .
pH+pOH=14
(d)
Expert Solution
Answer to Problem 20CR
pH of 5.65×10−5M Ca(OH)2 is 10.05 and pOH is 3.95.
Explanation of Solution
Concentration of OH- ion is 5.65×10−5M Ca(OH)2 in given solution. Here in calcium hydroxide there are two OH- groups therefore concentration of OH- will be obtained by multiplying given concentration by two. From this information pOH can be calculated.
These are in the wrong boxes. Why does the one on the left have a lower molar mass than the one on the right?
SYNTHESIS REACTIONS. For the following reactions, synthesize the given products from the given reactants.
Multiple reactions/steps will be needed. For the one of the steps (ie reactions) in each synthesis, write out the
mechanism for that reaction and draw an energy diagram showing the correct number of hills and valleys for
that step's mechanism.
CI
b.
a.
Use acetylene (ethyne)
and any alkyl halide as
your starting materials
Br
C.
d.
"OH
OH
III.
OH
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
(a) 0.200 M HCl
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell