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The thing to keep in mind here is that nitric acid is a strong acid, which means that it will ionize completely in aqueous solution to produce hydronium cations, h3o+, and nitrate anions, no− 3 Since molar mass is a fraction, g/mol, we can divide by multiplying by the reciprocal of the molar mass, mol/g. Hno3(aq) +h2o(l) → h3o+ (aq) + no− 3(aq) notice that every mole of nitric acid that is dissolved in water produces 1 mole of hydronium cations

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This means that the concentration of hydronium. Balanced equation mgo(s) + h_2o(l)rarrmg(oh)_2(s) moles magnesium hydroxide start with the given mass of mg(oh)_2 and convert it to moles by dividing by its molar mass (58.319 g/mol) 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 will be completely consumed by the reaction.

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. 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 This is also a 1:1 ratio. Conjugates are basically the other term

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). The longer the alkyl chain attached to the hydroxyl head, usually the more basic the conjugate base is (and the less nucleophilic). 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

You started with 0.1100 m hcl, but it was diluted from 40 ml to 100 ml Since water is in excess, 67.7 g mgo are needed to produce 98.0 g mg(oh)_2