Question

A student is given an antacid tablet that weighed 5.6832 g. The tablet was crushed and 4.3628 g of he antacid was added to 200 mL fo simulated stomah acid. This was allowed to react and the filtered. It was found that 25 mL fo this partially neutralized stomach acid required 8.5 mL of a NaOH solution to titrate it to a methyl red endpoint. If it took 25.5 mL of this NaOH solution to neutralize 25 mL of the original stomach acid.

a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?

b) how much stomach acid was neutralized y the 4.3628 g tablet?

c) how much stomach acid would have been neutralized by the original 5.6832 g tablet

Answer 1

(a)
`V_(neut)=8.33mL\ initial \ stomach \ acid`

(b)
`V_(4.3628gTablet)= 191.67 mL`

(c)
`V_(5.6832gTablet) = 249.67 mL\ acid`

Step-by-step explanation:

Hello,

(a) In this part, we need to consider that 25.5 mL of the NaOH solution is equivalent to 25.00 mL of the initial stomach acid . In such a way, the neutralized volume turns out:

`V_(neut)=8.5mLNaOH*(25mL\ initial \ stomach \ acid )/(25.5mLNaOH) \\V_(neut)=8.33mL\ initial \ stomach \ acid`

(b) Now, due to the fact that it takes 8.5 mL of NaOH to neutralize the previously computed 8.33 mL of the initial acid, the way to compute the neutralized volume by the 4.3628-g tablet is shown below:

`V_(4.3628gTablet)=200 mL- 8.33 mL = 191.67 mL`

(c) Finally, as long as 4.3628 g of the antacid is equivalent to 191.67 m.L of the initial acid , the way to calculate the neutralized volume by the 5.6832-g tablet is shown as follows:

`V_(5.6832gTablet)=5.6832 g\ antacid * (191.67 mL \  acid / 4.3628 g \ antacid) = 249.67 mL\ acid`

Best regards.

Answer 2

a) 8.33 ml of the original stomach acid is neutralized

b) 191.67 ml of the stomach acid was neutralized

c) 249.68 ml acid would be neutralized by the original tablet

Step-by-step explanation:

a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?

25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid

8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid

b) how much stomach acid was neutralized y the 4.3628 g tablet?

It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)

This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml

c) how much stomach acid would have been neutralized by the original 5.6832 g tablet

4.3628 g antacid is equivalent to 191.67 ml acid ( this is the answer for question 2)

5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid