Find pH of 0.1M NaOH solution.
A. 1
B. 2
C. 13
D. 17
Answer
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Hint: pH is a scale used to specify how acidic or basic a water-based solution is. Acidic solutions have a lower pH, while basic solutions have a higher pH.
Complete answer:
You will need to know the molarity of the NaOH. Let's assume the solution is 0.1M. NaOH is a strong base, so this will produce 0.1mol/L of OH ions in solution. This will produce a pH of 13.
pOH = -log \[\left[ {1 \times {{10}^{ - 1}}} \right]\]
\[ = - \log 1 + \left( { - \log {{10}^{ - 1}}} \right)\]
\[{\text{ = 0 + ( - ( - 1)) log 10}}\]
\[{\text{ = 0 + 1 x 1}}\]
\[{\text{ = 0 + 1}}\]
\[{\text{ = 1}}\]
Now we know that,
pH +pOH = 14
pH + 1 = 14
pH = 14 – 1
pH = 13
Therefore, the pH of 0.1 M NaOH solution is 13.
Note: The pH scale, which ranges from 0 to 14, tells you how acidic or alkaline a solution is. A pH lower than 7 is acidic, while a pH higher than 7 is alkaline. In mathematical terms, pH is the negative logarithm of the molar concentration of hydrogen ions in the solution. A pH testing strip will tell you that NaOH is a strong alkaline, but to calculate its exact pH, you have to work out its molarity first.
Complete answer:
You will need to know the molarity of the NaOH. Let's assume the solution is 0.1M. NaOH is a strong base, so this will produce 0.1mol/L of OH ions in solution. This will produce a pH of 13.
pOH = -log \[\left[ {1 \times {{10}^{ - 1}}} \right]\]
\[ = - \log 1 + \left( { - \log {{10}^{ - 1}}} \right)\]
\[{\text{ = 0 + ( - ( - 1)) log 10}}\]
\[{\text{ = 0 + 1 x 1}}\]
\[{\text{ = 0 + 1}}\]
\[{\text{ = 1}}\]
Now we know that,
pH +pOH = 14
pH + 1 = 14
pH = 14 – 1
pH = 13
Therefore, the pH of 0.1 M NaOH solution is 13.
Note: The pH scale, which ranges from 0 to 14, tells you how acidic or alkaline a solution is. A pH lower than 7 is acidic, while a pH higher than 7 is alkaline. In mathematical terms, pH is the negative logarithm of the molar concentration of hydrogen ions in the solution. A pH testing strip will tell you that NaOH is a strong alkaline, but to calculate its exact pH, you have to work out its molarity first.
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