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This is also a 1:1 ratio. You started with 0.1100 m hcl, but it was diluted from 40 ml to 100 ml 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. I got ph's of 1.36, 1.51, 1.74, 2.54 We want the standard enthalpy of formation for ca (oh)_2
Thus, our required equation is the equation where all the constituent elements combine to form the compound, i.e.
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 So this is a propanol derivative Both names seem to be unambiguous.
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
