Use the following experimental titration data to calculate the unknown concentra

Question

Use the following experimental titration data to calculate the unknown concentration and pH of the acid. Observation: The initial solution of acetic acid is clear and colourless. A few drops of phenolphthalein indicator are added to each sample. A dilute solution of potassium hydroxide is used as the titrant. As the mixture reaches the endpoint, flashes of pink colour are seen and the titrant is added drop by drop. The endpoint is reached when one drop of titrant turns the mixture a pale pink colour that does not fade.

Explanation / Answer

In the neutralization reaction one mole of hydromium (H+) reacts with one mole hudroxide (OH-). 10mL ofacetic acid reacts with 25.3 mL 10mL of acetic acid reacts with (50.51mL-25.3mL=25.21mL) 10mL of acetic cid reacts with (75.89-50.51mL=25.38mL) Volume(avg.)HC2H502=10mL Volume(avg.)KOH= (25.3mL+25.21mL+25.38mL)/3 =25.2967mL Macid=(Mbase)(Vbase)/Vacid Macid=7.589x10^-4; This would be the molar concentration of the acid (7.589x10^-4mo/L)(0.01L)=7.589x10^-6 mol acid (7.589x10^-6mol/L)(10mL/(25.2967mL+10mL… 2.15x10^-6 moles acid initial (3x10^-4mol/L)(25.2967mL/(25.2967mL+10m… 2.15x10^-4 moles base initial (2.15x10^-6moles)-(2.15x10^-4moles)= -2.129x10^-4: this means their is this much excess base times(-). Mbase=(2.129x10^-4moles)/(0.0252967L+0.… Kb=[OH-][S-]/[OHS], where S is the salt of the basw and OHS is the base. Then make ICE chart where I=initial concentration of substance in mol/L,C=the changes that inital concentrations undergoe during the reaction expressed by a variable determined from the stoichiometry of the equation, and E=The combiation of both, which we plug into to our Ka equation to solve for x. | KOH |(OH-) | K |; this comes from the Ka I| 6.03x10^-3 |0 | 6.0912x10^-5 | C| -x | x | x | E| | | | ----------------------------------------… chart is going to look a little messed due to the automatic syntax adjustment) (3x10^-4)=(6.03x10^-3-x)x/(x+6.0912x10^… plug in to the Kb equation and then solve for x simplifications: (6.0912x10^-5)/(3x10^-4)=20304, the hundred rule states if the answer to this calculation is greater than 100 than it's justifiable to drop the x since x is insignificant. so; (3x10^-4)=(6.03x10^-3-x)x/(6.0912x10^-5) 1.82736x10^-8=(6.03x10^-3)x -x^2 0=-x^2+(6.03x10^-3)x-(1.82736x10^-8); factorize x=3.032x10^-6 or x=6.027x10^-3 For this case, restrictions:x>6.0912x10^-5, since can't divide by zero x=3.032x10^-6 is inadmissable, which means x=6.027x10^-3 validate the earlier hypothesis by pluging in the x for the x that was removed as a result of applying the 100 rule. The answer is off by

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