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First, since NaOH is a base you have to find the pOH first so you use the equation -> pOH = -log[NaOH] pOH = -log[NaOH] = -log[0.0111] pOH = 1.955 Then you use this equation -> 14 = pH + pOH to find the pH 14 = pH + pOH pH = 14 - pOH = 14 - 1.955 pH = 12.045 and that makes it basic Hope that helped. ^_^
Supposed the acid and base are both strong: pH of the acid is 0.0 and the pH of the base (hydroxide) is 14.0
pH=10, so pOH=14-10=4 [OH]=10^-pOH [OH]=10^-4 [OH]=0.0001
pH and pOH are a measure of the concentration of the hydronium ions and hydroxyl ions respectively in the solution. pH = -log[H+] pOH = -log[OH-] and they are related: pH + pOH = 14
The pOH is 6,4.
First, since NaOH is a base you have to find the pOH first so you use the equation -> pOH = -log[NaOH] pOH = -log[NaOH] = -log[0.0111] pOH = 1.955 Then you use this equation -> 14 = pH + pOH to find the pH 14 = pH + pOH pH = 14 - pOH = 14 - 1.955 pH = 12.045 and that makes it basic Hope that helped. ^_^
1 millimolar = 0.001 M NaOH ( a base, remember ) - log(0.001 M NaOH) = 3 14 - 3 = 11 pH ----------
Supposed the acid and base are both strong: pH of the acid is 0.0 and the pH of the base (hydroxide) is 14.0
pH=10, so pOH=14-10=4 [OH]=10^-pOH [OH]=10^-4 [OH]=0.0001
pH and pOH are a measure of the concentration of the hydronium ions and hydroxyl ions respectively in the solution. pH = -log[H+] pOH = -log[OH-] and they are related: pH + pOH = 14
The pOH is 6,4.
The pOH is 8,7.
The pOH is 8,7.
The pOH is 8,7.
pH + pOH = 145.3 + pOH = 14pOH = 14 - 5.3pOH = 8.7
pOH +pH=14 pOH+7.6=14 pOH=(14-7.6)=6.4
The pH and pOH are related to each other through the equation: pH + pOH = 14 If the pH of a solution is 3, we can find the pOH by rearranging the above equation: pOH = 14 - pH = 14 - 3 = 11 Therefore, the pOH of the solution is 11.