That is a base.
It will have a pOH of 12. Because pH+pOH=14
First u need to know the pH of sulfuric acid: ph= -log[h+] =-log[0.4] 0.3979 now u can find out the POH of sulfuric acid ph+poh=14 poh=14-0.3979 poh=13.6
If a strong acid is mixed with a weak base, pH=pKa+/-1 in the buffer region.Corrected:If a strong base is mixed with a weak acid pOH= pKb +/-1 ( pH=(14 - pKb)+/-1) in the buffer region[Remember: For one conjugated pair of weak acid (a = HB) AND its weak base (b = B-):pKa + pKb = 14.0andpH + pOH = 14.0
pH is -log[H(subscript 3)O+] pOH is the [OH-] pOH = 14 - pH apex
p usually means the negative logarithim to base 10
Acid-base equations are solved by balancing the number of hydrogen ions (H+) and hydroxide ions (OH-) on both sides of the equation. This is done by identifying the acids and bases involved, determining their respective concentrations, and applying the principles of equilibrium and stoichiometry to calculate the pH or pOH of the solution. pH is calculated using the formula pH = -log[H+] and pOH is calculated using the formula pOH = -log[OH-].
pH and concentration of hydrogen ions (H+) are terms that refer to the amount of acid or base dissolved in a solution. pH is a measure of the acidity or alkalinity of a solution, while the concentration of hydrogen ions indicates the strength of an acid or base in a solution.
Yes, there is a pOH value in acidic solutions which is above 7.0 at the temperature of 298 K.
To calculate the pH of a weak base solution, you first need to determine the concentration of the base and the equilibrium constant (Kb) for the base's reaction with water. Then, use the equation pH 14 - pOH, where pOH is calculated using the concentration of the base and Kb. Finally, calculate the pH using the pOH value.
To find the pOH, take the negative logarithm of the hydroxide ion concentration. pOH = -log(OH-). Given [OH-] = 0.00015 M, pOH = -log(0.00015) ≈ 3.82.
The pOH can be determined by taking the negative base 10 logarithm of the hydroxide ion concentration. In this case, pOH = -log(0.0347) = 1.46.
For 104 the base is 10 and the exponent is 4.