The hydroxide ion concentration equals the hydronium ion concentration at a (neutral) pH of 7
The equation is acid + water equalizes into hydronium and conjugate base, and Ka (acid dissociation constant) is products divided by reactants. If the Acid = (H+)(base)/Ka, then the acid concentration is (H+)(H+)/Ka, or (0.0001)(0.0001)/0.0000001, which equals 1M.
Sodium hydroxide plus hydrochloric acid equals sodium chloride plus water.
sodium chloride and water
Aluminum hydroxide plus hydrochloric acid equals( produces ) aluminum chloride plus water.
4.0 x 10-11
The equation is acid + water equalizes into hydronium and conjugate base, and Ka (acid dissociation constant) is products divided by reactants. If the Acid = (H+)(base)/Ka, then the acid concentration is (H+)(H+)/Ka, or (0.0001)(0.0001)/0.0000001, which equals 1M.
This is a Bronsted question. Hs- is the acid in this which makes H2O a base. Therefore S-2 is the conjugate base and the H3O+ hydronium ion is the conjugate acid.
pH = -log(hydronium concentration) [Hydronium is H3O.-log(1 x 10-9) = 9
p = -log[H+] = 12.4
Yes, if both solutions are at 25oC then in both solutions this is valid: pH + pOH = 14.0
13.7 take the negative logarithm
Sodium hydroxide 1 N (normal solution) solution has a concentration of39,99710928 g/L (rounded 40 g/L) or 4 g/100 mL sodium hydroxide in water.
It depends on what the denominator was to start with: a surd or irrational or a complex number. You need to find the conjugate and multiply the numerator by this conjugate as well as the denominator by the conjugate. Since multiplication is by [conjugate over conjugate], which equals 1, the value is not affected. If a and b are rational numbers, then conjugate of sqrt(b) = sqrt(b) conjugate of a + sqrt(b) = a - sqrt(b), and conjugate of a + ib = a - ib where i is the imaginary square root of -1.
Sodium hydroxide plus hydrochloric acid equals sodium chloride plus water.
True
OH(-) A hydroxide. base
sodium chloride and water