Taking into account the definition of pH and pOH, the [OH⁻] is 7.08*10⁻¹⁰ M.
First of all, pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.
The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:
pH= - log [H⁺]= - log [H₃O⁺]
Being [H₃O⁺]=1.40×10⁻⁵ M, the pH is calculated as:
pH= - log (1.40×10⁻⁵ M)
Solving:
<u><em>pH= 4.85</em></u>
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Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
The following relationship can be established between pH and pOH:
pOH + pH= 14
Being pH= 4.85, pOH is calculated as:
pOH + 4.85= 14
pOH= 14 - 4.85
<u><em>pOH= 9.15</em></u>
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Replacing in the definition of pOH the concentration of OH⁻ ions is obtained:
- log [OH⁻]= 9.15
Solving
[OH⁻]= 10⁻⁹ ¹⁵
[OH⁻]= 7.08*10⁻¹⁰ M
In summary, the [OH⁻] is 7.08*10⁻¹⁰ M.
Learn more:
Answer : The length of protein will be, 36.8 A⁰
Explanation :
First we have to calculate the amino acid residue.

Given:
Length of single strand of protein = 27.0 kDa
Mean residue mass = 110 Da

Now we have to calculate the number of turns in protein.

As there are 3.6 amino acid per turn of the alpha-helix.

Now we have to calculate the length of the protein.

As, we know that the length of each turn = 0.54 A⁰

Thus, the length of protein will be, 36.8 A⁰
Answer:
if what you need is the final answer
the final answer is = 0.228 and I think the unit is Liter(L).
Explanation:
Answer:
The correct answer is 1.33 x 10⁻⁵ M
Explanation:
The concentration of the stock solution is: C= 1.33 M
In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:
C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M
The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
= 1.33 x 10⁻⁵ M
Why can’t y’all do more points then 5? why can’t it be like 20-10