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What is the pOH of ammonia?
The pOH of ammonia is approximately 4.7. Ammonia, NH3, is a weak base with a Kb value of approximately 1.8 x 10^-5. To find the pOH, you would first find the pH of the solution using the equilibrium constant for the base dissociation reaction and then use the relationship pOH = 14 - pH.
Calculate concentration of hydroxide ion with given pH?
Ist step, calculate pOH value by using formula pH + pOH = 14 2nd step, pOH = -log[OH], [OH] = - Antilog of pOH
A solution of HCl has H 2.5 x 10-5 What is the pOH of this solution?
pH + pOH = 14 If the pH is 3.4, the pOH is 10.6
What is the pH of a 10-5 m of koh solution?
The pH of a 10^-5 M KOH solution would be around 9. For a given concentration of a strong base like KOH, the pH can be calculated using the formula pH = 14 - pOH. Given that pOH = -log[OH-] and [OH-] = 10^-5 M in this case, pOH = 5. Therefore, pH = 14 - 5 = 9.
What is the pH and pOH of a 2.34x10-5 NaOH solution?
The pH of a 2.34x10^-5 NaOH solution is 12.33 (calculated as -log[OH^-]). The pOH of the same solution is 1.67 (calculated as -log[NaOH]).
What is the pOH of ammonia?
The pOH of ammonia is approximately 4.7. Ammonia, NH3, is a weak base with a Kb value of approximately 1.8 x 10^-5. To find the pOH, you would first find the pH of the solution using the equilibrium constant for the base dissociation reaction and then use the relationship pOH = 14 - pH.
Calculate concentration of hydroxide ion with given pH?
Ist step, calculate pOH value by using formula pH + pOH = 14 2nd step, pOH = -log[OH], [OH] = - Antilog of pOH
A solution of HCl has H 2.5 x 10-5 What is the pOH of this solution?
pH + pOH = 14 If the pH is 3.4, the pOH is 10.6
What is the pH of a solution with a hydroxide concentration OH of 1.0 x 10 12 M?
pH=-log[H+] pH=-log[1.00x10^-5M] pH=5 When concentration of H+ is 1M, the pH is zero. Every time the concentration is decreased by a magnitude of 10 (i.e. 10^1 ---> 10^-2) the pH goes up 1 value. This is true for the pH of bases as well, but in that case pH=14+log[OH-], which is derived from pOH=-log[OH] and pH+pOH=14. This is true for pure water at 25 degrees Celsius.
What is the pH of a 10-5 m of koh solution?
The pH of a 10^-5 M KOH solution would be around 9. For a given concentration of a strong base like KOH, the pH can be calculated using the formula pH = 14 - pOH. Given that pOH = -log[OH-] and [OH-] = 10^-5 M in this case, pOH = 5. Therefore, pH = 14 - 5 = 9.
If a solution had a pOH equals 10.7 what would be the H plus?
If an aqueous solution has a pOH value of 10.7 and is at standard temperature and pressure, the pH value is 14 - 10.7 = 3.3. From the definition of pH, this means that the logarithm (to base 10) of the molar concentration of H+, [H+] is -3.3. This can be written as +0.7 - 4. The antilog of 0.7 is 5, to the justified number of significant digits. Therefore, [H+] = 5 X 10-4.
What is the pH and pOH of a 2.34x10-5 NaOH solution?
The pH of a 2.34x10^-5 NaOH solution is 12.33 (calculated as -log[OH^-]). The pOH of the same solution is 1.67 (calculated as -log[NaOH]).
How do you calculate pOH?
Just like pH is the negative log of [H+], pOH is the negative log of the [OH-].
What is the pH of a solution with a hydroxide ion concentration of 3.7 x 10-5?
At 24C, the ionization product of water is 1.000 X 10-14. Therefore, the sum of pH and pOH is 14.0. pOHis 5 - 0.568 or about -4.33, and the pH must be 14.00 - 4.33 or 9.7, to the justified number of significant digits.
What is the pH if OH is 1x10 5 M?
- log(1 X 10^-5 M) = 5 14 - 5 = 9 pH ----------
What is the OH- of a solution with pH 5.75?
The hydroxide ion concentration of a solution with pH 5.75 can be calculated using the formula [OH-] = 10^(-pOH). First, find the pOH by subtracting the pH from 14 (pOH = 14 - pH = 14 - 5.75 = 8.25). Then, calculate [OH-] = 10^(-8.25) ≈ 5.62 x 10^(-9) mol/L.
What is the pH of a solution that has a (OH-) of 10-2?
pH=-log[H+] pH=-log[1.00x10^-5M] pH=5 When concentration of H+ is 1M, the pH is zero. Every time the concentration is decreased by a magnitude of 10 (i.e. 10^1 ---> 10^-2) the pH goes up 1 value. This is true for the pH of bases as well, but in that case pH=14+log[OH-], which is derived from pOH=-log[OH] and pH+pOH=14. This is true for pure water at 25 degrees Celsius.
