Acetic acid 1CH3COOH; Ka = 1.8 * 10-52 has a concentration in vinegar of 3.50% by mass. What is the pH of vinegar? (The density of vinegar is 1.02 g/mL.)

Video transcript

Hello. In this problem, we are told the concentration of hydrogen peroxide with an acid dissociation constant of 2.4 times 10 minus 12 in a bleaching agent solution is 6% by mass. Were asked to calculate the ph of the solution given that the density of the hydro peroxide solution is 1. g per milliliter. So begin. Then we're going to calculate our initial concentration of hydrogen peroxide. We can then make use of this in an ice table. So we are going to make an assumption that we have one liter of solution. So we have one leader of hydro peroxide solution. Will convert this two mL to one liter. It's equal to 1000 millionaires. And then we'll make use of the density 1.45 g. The hydrant box solution for every leader absolution. So our leaders of hydro peroxide solution cancels milliliters of solution cancels and we're left with a massive hydro peroxide solution. So you have 1450 g um solution. Within the problem statement, we're told that the% 6%. The show to buy mouse. This then tells us that we have six grams of hydrogen peroxide In every 100 g of the solution. So, using our massive solution found above and this percent by mass is a conversion factor set up to our massive solution cancels This works out to 87 g of hydrogen peroxide. Our initial concentrations of hydro peroxide is 87 g and one liter of solution and we'll use more massive hydro peroxide from Master moles. one mole of hydrogen peroxide has a massive 34.015g he said barr Mueller mass. So you know, it's a grams of peroxide cancel. And so we find the concentration that is 2.5577 solar. Next step then we'll make use of our initial concentration in a nice table. So hydrogen peroxide will act as a weak acid, so it will undergo hydraulic sis so it will react with water from the base of hydro peroxide and then joining mines. The hydro peroxide again acting as acid is going to donate a proton to water. So we have initial change and equilibrium. So initially our concentration of hydro peroxide then is 2. moller nor water since it's a pure liquid. And then our initial concentration of our products is zero, changes minus X plus X and plus X equilibrium. Then we can buy an initial and change. We have 2.5577 minus six X. And X. Their next step. Then we'll make use of our ask the dissociation constant to find our equilibrium concentration of joining mines. So our acid dissociation constant is equal to our product concentration over our reacting concentration and from the ice table, this works out to X times X over 2.5577 minus X. Which will be equal to then the K value of 2. Times 10 : -12. So we will given that we have a weak acid and a small acid dissociation constant. We're gonna check our simplifying assumption to see if we can then simplify this relationship and eliminate minus X. So to check our simplifying assumption, we're going to take our initial concentration of hydrogen peroxide divided by our acid dissociation constant. And this works out to 1.06, 6 times 10 to the 12, Which is much greater than 500. So we can then eliminate the minus X. And simplify our calculation. So then we get X squared Over 2.5577 is equal to 2.4 times 10 to the - and then find X. Move the 2.5577 to the other side and take the square root. And so x works out to 2.4776 times 10 to the -6. And this is equal to our joining my concentration and the last step then we're going to make you stand of our joining my concentration to calculate the ph so the ph is equal to the negative log of the adjoining mine concentration. So our ph works out to 5.61. So the page then of our solution is 5.61. This corresponds to answer d. Thanks for watching. Hope this help