pOH of Mixture of Two Strong Bases Solution

STEP 0: Pre-Calculation Summary
Formula Used
Negative Log of Hydroxyl Concentration = -log10((Normality of Solution 1*Volume of Solution 1+Normality of Solution 2*Volume of Solution 2)/(Volume of Solution 1+Volume of Solution 2))
pOH = -log10((N1*V1+N2*V2)/(V1+V2))
This formula uses 1 Functions, 5 Variables
Functions Used
log10 - The common logarithm, also known as the base-10 logarithm or the decimal logarithm, is a mathematical function that is the inverse of the exponential function., log10(Number)
Variables Used
Negative Log of Hydroxyl Concentration - The Negative Log of Hydroxyl Concentration gives us the value of hydroxyl concentration on the pH scale.
Normality of Solution 1 - (Measured in Equivalents per Liter) - The Normality of Solution 1 is described as the number of gram or mole equivalents of solution 1 present in one liter of solution 1.
Volume of Solution 1 - (Measured in Liter) - The Volume of Solution 1 gives the volume of solution 1 in liters.
Normality of Solution 2 - (Measured in Equivalents per Liter) - The Normality of Solution 2 is described as the number of gram or mole equivalents of solution 2 present in one liter of solution 2.
Volume of Solution 2 - (Measured in Liter) - The Volume of Solution 2 gives the volume of the solution 2 in liters.
STEP 1: Convert Input(s) to Base Unit
Normality of Solution 1: 0.0008 Equivalents per Liter --> 0.0008 Equivalents per Liter No Conversion Required
Volume of Solution 1: 0.00025 Liter --> 0.00025 Liter No Conversion Required
Normality of Solution 2: 0.0005 Equivalents per Liter --> 0.0005 Equivalents per Liter No Conversion Required
Volume of Solution 2: 0.0001 Liter --> 0.0001 Liter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
pOH = -log10((N1*V1+N2*V2)/(V1+V2)) --> -log10((0.0008*0.00025+0.0005*0.0001)/(0.00025+0.0001))
Evaluating ... ...
pOH = 3.14612803567824
STEP 3: Convert Result to Output's Unit
3.14612803567824 --> No Conversion Required
FINAL ANSWER
3.14612803567824 3.146128 <-- Negative Log of Hydroxyl Concentration
(Calculation completed in 00.004 seconds)

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Acidity and pH Scale Calculators

pH given Concentration of Hydrogen Ion
​ LaTeX ​ Go Negative Log of Hydronium Concentration = -log10(Concentration of Hydrogen Ion)
pH given Activity of Hydrogen Ion
​ LaTeX ​ Go Negative Log of Hydronium Concentration = -log10(Activity of Hydrogen Ion)
Concentration of Hydrogen Ion given pH
​ LaTeX ​ Go Concentration of Hydrogen Ion = 10^(-Negative Log of Hydronium Concentration)
Activity of Hydrogen Ion given pH
​ LaTeX ​ Go Activity of Hydrogen Ion = 10^(-Negative Log of Hydronium Concentration)

pOH of Mixture of Two Strong Bases Formula

​LaTeX ​Go
Negative Log of Hydroxyl Concentration = -log10((Normality of Solution 1*Volume of Solution 1+Normality of Solution 2*Volume of Solution 2)/(Volume of Solution 1+Volume of Solution 2))
pOH = -log10((N1*V1+N2*V2)/(V1+V2))

What is pOH scale?

pOH is a measure of hydroxide ion (OH-) concentration. It is used to express the alkalinity of a solution. Aqueous solutions at 25 degrees Celcius with pOH less than 7 are alkaline, pOH greater than 7 are acidic and pOH equal to 7 are neutral. pOH ranges from 1 to 14.

How to Calculate pOH of Mixture of Two Strong Bases?

pOH of Mixture of Two Strong Bases calculator uses Negative Log of Hydroxyl Concentration = -log10((Normality of Solution 1*Volume of Solution 1+Normality of Solution 2*Volume of Solution 2)/(Volume of Solution 1+Volume of Solution 2)) to calculate the Negative Log of Hydroxyl Concentration, The pOH of Mixture of Two Strong Bases formula is defined as the negative base-10 logarithm of the concentration of hydroxyl ion of a solution which is calculated using normality and volume of two strong acids of 1 and 2. Negative Log of Hydroxyl Concentration is denoted by pOH symbol.

How to calculate pOH of Mixture of Two Strong Bases using this online calculator? To use this online calculator for pOH of Mixture of Two Strong Bases, enter Normality of Solution 1 (N1), Volume of Solution 1 (V1), Normality of Solution 2 (N2) & Volume of Solution 2 (V2) and hit the calculate button. Here is how the pOH of Mixture of Two Strong Bases calculation can be explained with given input values -> 3.103219 = -log10((0.8*2.5E-07+0.5*1E-07)/(2.5E-07+1E-07)).

FAQ

What is pOH of Mixture of Two Strong Bases?
The pOH of Mixture of Two Strong Bases formula is defined as the negative base-10 logarithm of the concentration of hydroxyl ion of a solution which is calculated using normality and volume of two strong acids of 1 and 2 and is represented as pOH = -log10((N1*V1+N2*V2)/(V1+V2)) or Negative Log of Hydroxyl Concentration = -log10((Normality of Solution 1*Volume of Solution 1+Normality of Solution 2*Volume of Solution 2)/(Volume of Solution 1+Volume of Solution 2)). The Normality of Solution 1 is described as the number of gram or mole equivalents of solution 1 present in one liter of solution 1, The Volume of Solution 1 gives the volume of solution 1 in liters, The Normality of Solution 2 is described as the number of gram or mole equivalents of solution 2 present in one liter of solution 2 & The Volume of Solution 2 gives the volume of the solution 2 in liters.
How to calculate pOH of Mixture of Two Strong Bases?
The pOH of Mixture of Two Strong Bases formula is defined as the negative base-10 logarithm of the concentration of hydroxyl ion of a solution which is calculated using normality and volume of two strong acids of 1 and 2 is calculated using Negative Log of Hydroxyl Concentration = -log10((Normality of Solution 1*Volume of Solution 1+Normality of Solution 2*Volume of Solution 2)/(Volume of Solution 1+Volume of Solution 2)). To calculate pOH of Mixture of Two Strong Bases, you need Normality of Solution 1 (N1), Volume of Solution 1 (V1), Normality of Solution 2 (N2) & Volume of Solution 2 (V2). With our tool, you need to enter the respective value for Normality of Solution 1, Volume of Solution 1, Normality of Solution 2 & Volume of Solution 2 and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Negative Log of Hydroxyl Concentration?
In this formula, Negative Log of Hydroxyl Concentration uses Normality of Solution 1, Volume of Solution 1, Normality of Solution 2 & Volume of Solution 2. We can use 2 other way(s) to calculate the same, which is/are as follows -
  • Negative Log of Hydroxyl Concentration = -log10(Concentration of Hydroxyl Ion)
  • Negative Log of Hydroxyl Concentration = 14+log10((Normality of Solution 1*Volume of Solution 1-Normality of Solution 2*Volume of Solution 2)/(Volume of Solution 1+Volume of Solution 2))
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