[OH−] = 10−4M This goes to show that an acid was present in the solution, since the concentration of hydroxide anions is smaller than it would have been for the magnesium …

Explanation: Your starting point here is the pH of the solution. More specifically, you need to use the given pH to determine the concentration of hydroxide anions, #"OH"^ (-)#, present in the …

We want the standard enthalpy of formation for Ca (OH)_2. Thus, our required equation is the equation where all the constituent elements combine to form the compound, i.e.: Ca …

The degree of dissociation sf (alpha=0.0158) sf (K_b=2.51xx10^ (-6)color (white) (x)"mol/l") Triethyamine is a weak base and ionises: sf ( (CH_3)_3N+H_2Orightleftharpoons …

Here's what I got. The problem wants you to use the base dissociation constant, K_b, of ammonia, "NH"_3, to determine the percent of ammonia molecules that ionize to produce …

Since water is in excess, "67.7 g MgO" are needed to produce "98.0 g Mg(OH)"_2. Balanced equation "MgO(s) + H"_2"O(l)"rarr"Mg(OH)"_2("s")" Moles magnesium hydroxide Start with the …

pH = 1.61151 OH^- = 4.08797 * 10 ^-13M HF = 0.855538M H^+ = 0.024462M F^- = 0.024462M HF + H_2O = H_3O^+ + F^- We can find the concentration of H^+ or H_3O^+ by three ways …

Question 1: K_ (sp)= 1.1 xx10^ (-11) Question 2: s= 4.9 xx10^ (-12)M Quest (1) determine the ksp for magnesium hydroxide Mg (OH)_2 where the molar solubility of Mg ...

The sodium ions remain in solution as spectator ions. If XS sodium hydroxide is added the precipitate redissolves to give the soluble plumbate (II) ion. A simple way of writing this is: …

CH 3COOH (aq) +OH − (aq) → CH 3COO− (aq) +H 2O(l) Notice the 1:1 mole ratio that exists between acetic acid and sodium hydroxide (written as hydroxide ions). This means that, in …