Question
Could you please help me solving these questions by showing me the calculations,,
hen a solution of a strong acid is mixed with a solution of a strong base, a chemical reaction occu that can be represented by the following net ionic equation: H'(aq) + OH-(aq) H2O(1) and chemists use it extensively to change the acidic or basic proper- This is called a neutralization reaction ties of solutions. The equilibrium constant for th tion can be considered to proceed completely to the right, using up whichever of the ions is present in the lesser amount and leaving the solution either acidic or basic, depending on whether H or OH ion was in utions. The equilibrium constant for the reaction is about 10'4 at room temperature, so that the reac excess. ince the reaction is essentially quantitative, it can be used to determine the concentrations of acidic or basic solutions. A frequently used procedure involves the titration of an acid with a base. In the titration a basi c solution is added from a buret to a measured volume of acid solution until the number of moles o OH ion added is just equal to the number of moles of Ht ion present in the acid. At that point the volume of basic solution that has been added is measured Recalling the definition of the molarity, MAx of species A, we have moles of A A-liters ofsolution_ moles of A-M, X V or At the end point of the titration moles H+ originally present = moles OH-added So, by Equation 1, solution is known, the molarity of the other ion can be found from the titration. The equivalence point or end point in the titration is determined by using a chemical, called an indicator that changes color at the proper point. The indicators used in acid-base titrations are weak organic acid bases that change color when they are neutralized. which is colorless in acid solutions but becomes red when the pH of the solution becomes 9 or higher s or One of the most common indicators is phenolphthalein strong acid is titrated with a solution of a strong base, the pH at the end point will be When a solution of a about 7. At the end point a drop of acid or base added to the solution will change its pH by several pH units so that phenolphthalein can be used as an indicator in such titrations. If a weak acid is titrated with a base, the pH at the equivalence point is somewhat higher than 7, perhaps 8 or 9, and phenolphthalein is still a very satisfactory indicator. If, however, a solution of a weak base such as ammonia is titrated with a acid, the pH will be a unit or two less than 7 at the end point, and phenolphthalein will not be as good an cator for that titration as, for example, methyl red, whose color changes from red from about 5 to 6. Ordinarily, indicators will be chosen so that their color change occurs at about the pH at equivalence point of a given acid-base titration. weak acid is titrated with a stron ever, a solution of a weak base such as ammonia is titrated with a stron n indi to yellow as the pH changes ll determine the molarity of OH In this experiment you will determine the molarity of OH ion in an NaOH solution tion against a standardized solution of HCI. Since in these solutions one mole of acid in solution ion in an NaOH solution by titrating that solu in solution furnishes one e mole of acid
Explanation / Answer
1) a)
we know that
moles = molarity x volume (L)
so
moles of NaOH = 6 x 7.2 x 10-3
moles of NaoH = 0.0432
b)
now given
final volume = 400 ml = 0.4 L
so
molarity = moles / volume (L)
so
molarity of NaOH = 0.0432 / 0.4
molarity of NaOH = 0.108
so
the molarity of NaOH is 0.108 M
2)
a)
we know that
HCl ---> H+ + Cl-
so
Molarity of HCl = molairty of H+ = 0.0997 M
b)
now
MH+ X Vacid = MOH- x Vbase
so
using given values
we get
0.0997 x 20.04 = MOH- x 21.16
MOh- = 0.0944
so
MOH- in NaOH solution is 0.0944 M
c)
now
NaOH --> Na+ + OH-
so
MNaOH = MOH- = 0.0944 M
3)
we know that
MOH- = MNaOH = 0.0944
now
moles of OH- = MOH- x volume (L)
moles of OH- = 0.0944 x 24.13 x 10-3
moles of OH- = 2.2779 x 10-3
moles of OH- used = 2.2779 x 10-3
now
moles of H+ from HCl = MHcl x V
moles of h+ from HCl = 0.0997 x 0.32
moles of H+ from HCl = 3.1904 x 10-5
b)
now
moles of H+ in solid acid = moles of OH- in NaOH - moles of H+ in HCl
so
moles of H+ in solid acid = 2.2779 x 10-3 - 3.1904 x 10-5
moles of H+ in solid acid = 2.246 x 10-3
so
moles of H+ in solid acid = 2.246 x 10-3
c)
now
molar mass of unknown acid = grams of acid / moles of H+
so
molar mass = 0.3012 / 2.246 x 10-3
molar mass = 134
so
the molar mass of unknown acid is 134 g/mol