What is the pH of a 0.020 M NaOH solution

NaOH is a strong base, meaning it will dissociate basically completely in solution to form #Na^+# and #OH^-# ions in solution.

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The correct option is B 12.3Since sodium hydroxide is a strong base, it will dissociate completely in water. This means that the concentration of the base will be equal to the concentration of hydroxide ions after the reaction runs to completion. NaOH(aq) ⇌Na++OH−0.02M 0.02M Since 1 hydroxide ion is created from each molecule of sodium hydroxide that dissociates , hence 0.02 M NaOH will give 0.02 M of hydroxide ions. pOH=−log10(OH)−=−log10(0.02)pOH=1.70pH+pOH=14⇒pH=14−1.70=12.3pH=12.3
What is the pH of a 0.02 M NaOH solution?
What is the pH of a 0.020 M?
What is the pH of a 0.0500 m NaOH solution?
What is the pH of a 0.0125 m NaOH solution?

One mole of #OH^-# ions is created for every mole of NaOH that is dissolved. If it's a .05 M solution of NaOH, then it can also be interpreted to be a .05M solution of #OH^-#.

#pOH# (The pH scale for bases) can be found by taking the negative logarithm of #OH^-# concentration.

#pOH# = #-log[.05 M]# = 1.3.

However, we are trying to find #pH#, so we must use the formula
#pOH + pH = 14#.

Now that we know #pOH = 1.3#,
#pH = 14 - 1.3 = 12.7#.

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Solution

The correct option is B 12.3Since sodium hydroxide is a strong base, it will dissociate completely in water. This means that the concentration of the base will be equal to the concentration of hydroxide ions after the reaction runs to completion. NaOH(aq) ⇌Na++OH−0.02M 0.02M Since 1 hydroxide ion is created from each molecule of sodium hydroxide that dissociates , hence 0.02 M NaOH will give 0.02 M of hydroxide ions. pOH=−log10(OH)−=−log10(0.02)pOH=1.70pH+pOH=14⇒pH=14−1.70=12.3pH=12.3

What is the pH of a 0.02 M NaOH solution?

pOH=−log10(OH)−=−log10(0.02)pOH=1.70pH+pOH=14⇒pH=14−1.70=12.3pH=12.3.

What is the pH of a 0.020 M?

For example: What is the pH of a 0.020 M HCl solution? Answer: HCl is a strong acid, it dissociates 100% (I.e. completely): [HCl]F = 0.020 M = [H3O+] So, pH = - log [H3O+] = -log [HCl]F = -log(. 020)=1.70 Voila!