Answer: 7
Explanation:
50 ml 0.125 M KOH = (50 * 0.125) = 6.25 ml 1 M KOH
similarly 50 ml 0.125 M HCl = 6.25 ml 1 M HCl
so KOH will fully neutralize the HCl so the pH of the medium will be 7
H+ + OH- = H2O
so Kw = [H=][OH-]
= [H+]^2 = 10-14
SO [H+] = 10-7
pH = -log [H+]
pH = -log[10-7]
pH = 7
Answer:
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
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
Answer:
These reagents are also called as reductants. Redox reactions: Reactions which involve change in oxidation number of the interacting species. Problem 8.4 Justify that the reaction: 2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g) is a redox reaction.
