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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 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. 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
This means that the concentration of hydronium. Since molar mass is a fraction, g/mol, we can divide by multiplying by the reciprocal of the molar mass, mol/g. 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
The longer the alkyl chain attached to the hydroxyl head, usually the more basic the conjugate base is (and the less nucleophilic). 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 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
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)
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