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0-/0.1 points0/4 Submissions Usedbuffer solution contalns 37 mol of hydrocyanlc acld (HCN) ad 0.58 mol of sodlum cyanlde (NaCN) In 8.00 The Ka of hydrocyanlc acid (...

Question

0-/0.1 points0/4 Submissions Usedbuffer solution contalns 37 mol of hydrocyanlc acld (HCN) ad 0.58 mol of sodlum cyanlde (NaCN) In 8.00 The Ka of hydrocyanlc acid (HCN) Is Ka 9e-10.(a) What Is the pH of thls buffer?(b) What the pH of the buffer after the additlon of 0.32 mol of NaOH? (assume no volume change)(c) What is the pH of the origina buffer after the additlon of 0.13 mol of HI? (assume no volume change)

0-/0.1 points 0/4 Submissions Used buffer solution contalns 37 mol of hydrocyanlc acld (HCN) ad 0.58 mol of sodlum cyanlde (NaCN) In 8.00 The Ka of hydrocyanlc acid (HCN) Is Ka 9e-10. (a) What Is the pH of thls buffer? (b) What the pH of the buffer after the additlon of 0.32 mol of NaOH? (assume no volume change) (c) What is the pH of the origina buffer after the additlon of 0.13 mol of HI? (assume no volume change)



Answers

A buffer solution contains $0.10 \mathrm{~mol}$ of propionic acid $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)$ and $0.13$ mol of sodium propionate $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COONa}\right)$ in $1.50 \mathrm{~L} .$ (a) What is the $\mathrm{pH}$ of this buffer? (b) What is the pH of the buffer after the addition of $0.01$ mol of $\mathrm{NaOH} ?$ (c) What is the $\mathrm{pH}$ of the buffer after the addition of $0.01 \mathrm{~mol}$ of HI?

Here is another Henderson Hasselbach equation. Just in this problem, we are given the following information. We were asked to calculate the pH of a solution. Let me see if I've got this going as to calculate the pH of a solution that is 0.40 Moeller h two n n. H. Two. This solution is also 0.80 Mueller h two, n n h. Three and all three. The first substance is a weak base, and the nitrate ion substance is its conjugate acid. In order for this buffer to have a pH that was equal to the peak A. Would you add HCL or Noh? So would you add acid or would you add base? And also what quantity of which re agent would you add to one leader of the original buffer so that the resulting solution had a PH equal to peek? A. So the P K B is equal to 3.0 times 10 to the minus six, so the P K is equal to the K W. One times 10 to the minus 14 minus K B. And there's the answer for this. 333 times 10 to the minus ninth. So the P H is equal to the negative log of 3.33 Well, let's just right K plus the log of the base over the concentration of the asset. So P H is equal Thio negative log of 3.33 times 10 to the minus ninth. Plus the log of 0.40 is your base and 0.80 is our asset. So the ph is 8.18 That was the first part of what we needed to solve. Boy, that page doing that was weird. Test to Now we're going to figure out if we're gonna add HCL or a H in order to bring our make her pH equal to R P. A. So, in order for the pH to be equal to P. K, the log of the base over the lab of the log of the acid, that ratio has to equal one because the log of one is zero. So if the base concentration equals the acid concentration, well, we have zero. So we since we have played 40 Moeller of the base used to be, I'm yawning. You could probably hear me doing that? I was wondering who's going with that. So the base over the acid, 0.40 0.80 So we're gonna have to add 0.20 bulls per leader. If we have a leaders solution of the H that would give us 0.6 over 0.6. So we'll add Played 20 moles per leader of any Ohh, and that should be the answer.

To calculate the pH of a solution that contains sodium cyanide and hydro scion IQ acid. We simply need to use the Henderson Astle Mulch equation where pH is equal to P. K. A. PK is given to us at 9.31 plus the log of the moles of the base. The base in this case is cyanide. So to calculate, the molds will take the volume, which is 80 milliliters, or 800.800 leaders multiplied by the concentration, which is 0.3 Moeller. This will give us the moles of the base within divide by the moles of the asset, which is the 20 mil leaders multiplied by the 200.5 Moller concentration, and we get a pH of 9.69 for the 2nd 1 to calculate. The pH of the buffer solution will also use to Anderson Hassle Baulch equation. Although the PK A for pipes is not given to us, so we need to look it up. In the textbook, it's 6.8, so pH is equal to p K a plus the log of the moles of the base. The base in this case is the 60 mil leaders multiplied by the 600.15 Moeller. And then the moles of the acid form of pipes will be the 40 milliliters multiplied by the 400.30 Moeller, and we get a pH of 6.68

First, we'll calculate the pH of the buffer solution. To do that, we need to know the K A of the conjugate acid to h two n N. H. Two. We can determine this by looking up. The K B, which is 3.0 times, tend the negative 64 h two and H two Divide that indicate W. And we'll get the K of 3.33 times 10 the negative nine. So the pH of the buffer solution will be equal to the negative log of the K value, which is P. K. Plus the concentration of the base at point forward, divided by the concentration of the asset. At 0.0.8, we get a pH of 8.18 when pH equals P K A. This ratio right here equals one. In order to make that ratio one, we need to increase the amount of base and decrease the amount of acid. The only way to do that is to add a strong base such as sodium hydroxide. Then to figure out the moles of strong base, we need to add will set the ratio equal toe one, which is what is desired and have the base amount be 0.4, plus the amount of strong base added and the weak acid amount be 0.8, minus the amount of strong base added and then solving for the moles of base X, we get point to, so we would add 0.2 moles of strong base sodium hydroxide.


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