The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide according to: CH3COOH1aq2 + NaOH1aq2¡ H2O1l2 + NaCH3COO1aq2 If 3.45 mL of vinegar needs 42.5 mL of 0.115 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00-qt sample of this vinegar?

Video transcript

Hi everyone here we have a question telling us the equation for the reaction between nitrous acid and potassium hydroxide is nitrous acid plus potassium hydroxide forms potassium nitrite plus water When 18.92 ml of Nitrous acid was tight traded with 0. molar potassium hydroxide, 19.52 ml of potassium hydroxide was required towards the equivalent point. And our goal is to calculate the mass of nitrous acid ingram's present in a one quart sample. So we're going to start off with our zero .156 malls a potassium hydroxide and that is per leader because polarity is most per leader And we want to get that in just mold. So we're going to multiply that by our 0. L. And we're going to multiply that by our multiple ratio. So one mole of nitrous acid over one mole of potassium hydroxide. And we're going to multiply that by The molar mass of nitrous acid. So .013g of nitrous acid over one mole of nitrous acid. So our moles of potassium hydroxide are canceling out, our leaders are canceling out are moles of nitrous acid are canceling out leaving us with grams of nitrous acid and that equals 0. 16 grams of nitrous acid and that is in 18 . ml. Now we're going to calculate for one court. So we have one court and we're going to multiply that by one leader Over 1.057 quarts. And we're going to multiply that by 1000 ml over one leader. And we're going to multiply that by the g we just calculated so zero 14 g of nitrous acid. Over Our 18. ml. And our courts are canceling out. Our leaders are canceling out and our middle leaders are canceling out, Leaving us with 7.1586 g of nitrous acid per court. And that is our final answer. Thank you for watching. Bye.