A 5.55-g sample of a weak acid with Ka = 1.3 * 10 - 4 was combine...

Question

A 5.55-g sample of a weak acid with Ka = 1.3 * 10 - 4 was combined 
with 5.00 mL of 6.00 M NaOH, and the resulting solution 
was diluted to 750.0 mL. The measured pH of the solution was 
4.25. What is the molar mass of the weak acid?

Accepted Answer

welcome back everyone in this example, we're finding the molar mass of an unknown weak acid. In the given scenario, we have a 2.5 mL solution of 3.0 molar sodium hydroxide mixed with a 6.51 g sample of an unknown weak acid with a K. A. Of 6.2 times 10 to the 10th power. We're told that the resulting solution is diluted to 500 mL and is measured to have a ph of 7.72. So our first step because we're dealing with an acid and a base is to write out our equilibrium reaction. So we have according to the prompt sodium hydroxide as our strong base reacting with our weak acid, which is unknown. So we're going to use a general symbol for a weak acid being H A. Now we have our equilibrium symbol and we should recall that are weak asset is going to lose a proton where that proton is going to form a bond with hydroxide to form our product being water as well as our conjugate base represented by a minus here. And so we now want to recognize that this equation is actually balanced as we have written it out. And this is important because we recognize that we have a mono protic acid meaning we just have one hydrogen proton here. And because this is a mono protic acid, that means that we're going to get one mole of our conjugate base for every mole of hydroxide that reacts and because we see that we have one mole of hydroxide reacting. That means that our moles of our sodium hydroxide are equivalent to our moles of our conjugate base. So we have our moles of sodium hydroxide That should equal our moles of our conjugate base. And so we're going to find our moles of sodium hydroxide first by using the info from the prompt. So we're given our volume of sodium hydroxide as 2.50 ml. And we're also given our polarity of sodium hydroxide as 3. Moller. But we want to recall that polarity Is equivalent to moles divided by leaders. And so this is interpreted as 3.0 moles divided by leaders. And because we have leaders in the denominator that means that our volume of our sodium hydroxide needs to be converted to leaders. So we're going to make that conversion here. We have leaders as our final unit and middle leaders in the denominator. So it can cancel out. And we should recall that our prefix milli tells us that we have 10 to the negative third power of our base unit leader for one middle leader. This allows us to cancel our units of middle leaders leaving us with leaders as our final unit. And this is going to give us a volume of 0. liters of our sodium hydroxide. And now that we have everything in the proper units, we're going to take that volume that we just found 0. liters of sodium hydroxide. And we're going to multiply it by our polarity which we understand as three point oh oh moles divided by leaders. This allows us to cancel out leaders leaving us with moles as our final unit. And this is going to give us our moles of sodium hydroxide equal to the value of 7.5 times 10 to the negative third power moles. And this we can understand is also equal to our moles of our are conjugate base here. So we can say therefore are moles if our conjugate base is also equal to 7.5 times 10 to the negative third power moles. Now that we understand our moles of our conjugate base. We want to recall our formula for ph and we should recall that ph can be calculated by taking our P. K. A. Of our weak acid and adding that to the log of the quotient. Where we take our moles of our conjugate base divided by our moles of our weak acid. And so in this case we'll just say divided by moles of H. A. And so because we are given from the prompt R. K. A. Of our weak acid, we want to convert R K. A. To R P K. So we're going to recall that R P K. A. It's equal to we would say the negative log of R K. A. And so we're going to get that. We have the negative log of our given K. From the prompt as 6.2 times 10 to the negative 10th power. And this is going to give us our PK of 9.21 for our weak acid. So let's plug that into our formula below. So beginning with our ph from the prompt, were given a ph of 7.72. We're setting this equal to R P K. Which we just determined to be the value 9.21. This is then added to the log of our moles of our conjugate base. So we have the value we calculated being 7.5 times 10 to the negative third power moles of our conjugate base divided by our moles of our weak acid which we are solving for here. And so we need to simplify this by first subtracting from both sides 9.21. So this cancels this out. And what we're going to get for a difference is on the left hand side, a value of negative 1.49 which is set equal to our right hand side where we have the log of our quotient. So we have our moles of our We have our moles of our conjugate base 7.5 times 10 to the negative third power moles divided by again our moles of our weak acid. And so to further simplify this, we're going to make each side an exponents to the base of 10. So we have 10 here and 10 here. This cancels this log term on the right hand side. And what the left hand side simplifies to is the value 0.3235937. This is set equal to the right hand side, where we have our quotient, where we have 7.5 times 10 to the negative third power moles of our conjugate base divided by our moles of our weak acid. And now to further isolate this or simplify this. We want to recognize that we have a diagonal when we have diagonals in algebra we can trade places. And so we would say that our moles of our weak acid is equal to the quotient, where we have 7.5 times 10 to the negative third power moles of our conjugate base divided by 0.03235937. This gives us our moles of our weak acid equal to a value of 0. moles. So we finally have our moles of our weak acid. And we want to recall that our final answer should be our molar mass. And recall that molar mass is units of g per mole. And so going back to the prompt, we can see that we're given our mass of our weak acid being a 6.51 g sample. So we're gonna plug that in. And what we'll say is our molar mass of our weak acid is equal to our grams of our weak acid given in the prompt as 6. g divided by our moles of our weak acid, which we determined above as zero point moles. And this quotient here gives us a final results as a molar mass equal to a value of about 28.1 g per mole. Now we want to recognize that our smallest number of sig figs from the prompt is r k a given as 6.2. And so we're going to round this 2 to 66 and say that this is about 28 g per mole. And so now we have our final answer being our molar mass of our weak acid. So what's highlighted in yellow completes this example as our final answer. If you have any questions, leave them down below and I'll see everyone in the next practice video

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