H^+ + OH^--> H_2O when the acid was added to the resulting solution. The H^+ and OH^- react in a 1:1 ratio. This tells us that the number of moles of H^+ used will be equal to the number of …

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 …

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 acid in excess is then titrated with N aOH (aq) of KNOWN concentration....we can thus get back to the concentration or molar quantity of M (OH)2...as it stands the question (and answer) …

Similarly, OH^- becomes H_2O, indicating a gain of a H^+ ion. So, you can say that NH_4^+ is the acid, and OH^- is the base. Conjugates are basically the "other" term. For every acid, you …

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: …

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 …

Explanation: #H_3PO_4 (aq)+Ca (OH)_2 (aq) rarr Ca_3 (PO_4)_2+H_2O (l)#