"6.3072 g" >>"Molarity" = "Moles of solute"/"Volume of solution (in litres)" "0.45 M" = "n"/"0.4 L" "n = 0.45 M × 0.4 L = 0.18 mol" You need "0.18 mol" of "NH"_4"OH" Molar mass of "NH"_4"OH" …

The effect of strong base on water is to dramatically increase the concentration of OH^- ions and decrease the concentration of H_3O^+ ions. Water always contains at least small …

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 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) …

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 …

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 …

OH− (aq) + H3O+ (aq) → 2H2O(l) so you can say that when you mix these two solutions, the hydronium cations present in the hydrochloric acid solution will be the limiting reagent, i.e. they …

The added water to reach "100.00 mL" doesn't change the mols of HCl present, but it does decrease the concentration by a factor of 100//40 = 2.5. Regardless, what matters for …

calculate moles of Acid, calculate moles of Base. subtract, #OH^-# should be in excess, calculate Molarity of #OH^-# , pOH = -log (#OH^-#), then 14-pOH = pH

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 …