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This is also a 1:1 ratio. 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. 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
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 Since molar mass is a fraction, g/mol, we can divide by multiplying by the reciprocal of the molar mass, mol/g. This means that the concentration of hydronium.
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). 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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