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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) are hypothetical. 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 One is by the ICE table (but this is a 5% rule) and the other is square root which is absolutely correct and the other is Ostwald's law of dillution Let's set up an ICE table. color (white) (mmmmmmmm)"HF" + "H"_2"O" ⇌ "H. 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

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They will be completely consumed by the reaction. You started with 0.1100 m hcl, but it was diluted from 40 ml to 100 ml > (a) with hcn the hcn adds across the α c=o group to form a cyanohydrin

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Underbrace (ch_3cocooh)_color (red) (pyruvic acid) + hcn →.

So this is a propanol derivative Both names seem to be unambiguous. 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_4oh molar mass of nh_4oh is 35.04 g/mol mass of solute = 0.18 cancelmol × 35.04 g/cancelmol = 6.3072 g Start by writing the balanced chemical equation for this neutralization reaction

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 neutralization is what amount of naoh you add to what number of mols of hcl I got ph's of 1.36, 1.51, 1.74, 2.54