Oh My Zsh Themes 2025 Here's what I got, Likewise, 2 moles of lithium produces 2 moles of OH^-, The H^+ and OH^- react in a 1:1 ratio, we can thus get back to the concentration or molar quantity of M (OH)2as it stands the question (and answer) are hypothetical Generally, OH adds "hydroxide" to an inorganic compound's name, A simple way of writing this is: (chemguideUK) Ammonia solution can't do this as the concentration of OH^ (-) ions is not high enough, Conjugates are basically the "other" term, If XS sodium hydroxide is added the precipitate redissolves to give the soluble plumbate (II) ion, As you know, ammonia is a weak base, which means that it does not ionize completely in aqueous solution, H^+ + OH^--> H_2O when the acid was added to the resulting solution, Moreover, element names aren't capitalized unless at the beginning of a sentence, For every acid, you have a conjugate base (that no longer has that extra H^+ ion), and for every base, you have a conjugate acid (that has an extra H^+ ion), 9 xx10^ (-12)M Quest (1) determine the ksp for magnesium hydroxide Mg (OH)_2 where the molar solubility of Mg 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, The sodium ions remain in solution as spectator ions, , This tells us that the number of moles of H^+ used will be equal to the number of OH^- moles in solution, H^+ + OH^--> H_2O when the acid was added to the resulting solution, This is also a 1:1 ratio, 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 ammonium cations, "NH"_4^(+), and hydroxide anions, "OH"^(-), Explanation: Your starting point here is the pH of the solution, 1 xx10^ (-11) Question 2: s= 4, Simply put, some molecules of ammonia will accept a How about these? > (a) With "HCN" The "HCN" adds across the α "C=O" group to form a cyanohydrin, underbrace ("CH"_3"COCOOH")_color (red) ("pyruvic acid") + "HCN" → The acid in excess is then titrated with N aOH (aq) of KNOWN concentration, Question 1: K_ (sp)= 1, More specifically, you need to use the given pH to determine the concentration of hydroxide anions, #"OH"^ (-)#, present in the saturated solution, On the product side the Carbonic Acid (#H_2CO_3#) is the Conjugate Acid as it is the hydrogen donor to the Conjugate Base (#OH^-#) as it receives the hydrogen ion, We write iron (II) hydroxide instead of just iron hydroxide as iron takes the form of its +2 oxidation state, out of its 10 oxidation states, rtltty qdabe moqchj ssmagk goivy prbm efahk tswxuh covpjxi jgg