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Calculate the pH of = buffer system that contains 0.50 M of NH; (aq) ad 0.15 M of NH,Cl(aq) Note that the Kh value of NH; (aq) Is 1.8 x 10pH:b) Determine the change...

Question

Calculate the pH of = buffer system that contains 0.50 M of NH; (aq) ad 0.15 M of NH,Cl(aq) Note that the Kh value of NH; (aq) Is 1.8 x 10pH:b) Determine the change In pH if 2.00 mL of 0.100 M HCl is added to 0.040 L of the buffer system described In part _ApH:c) Determine the change in pH if 2.00 mL of 0.100 M NaOH is added to 0.040 L of the buffer system described in part a)ApH:

Calculate the pH of = buffer system that contains 0.50 M of NH; (aq) ad 0.15 M of NH,Cl(aq) Note that the Kh value of NH; (aq) Is 1.8 x 10 pH: b) Determine the change In pH if 2.00 mL of 0.100 M HCl is added to 0.040 L of the buffer system described In part _ ApH: c) Determine the change in pH if 2.00 mL of 0.100 M NaOH is added to 0.040 L of the buffer system described in part a) ApH:



Answers

Calculate the $\mathrm{pH}$ of the buffer system made up of $0.15 M \mathrm{NH}_{2} / 0.35 M \mathrm{NH}_{4} \mathrm{Cl}$

Here's another Henderson Hasselbach equation, and this is another voice over for this equation. Were given some information and we were asked to find the pH of a solution. And the solution is 0.20 Mueller H L C L, and were asked to find how many moles of either an acid or base so that we can guarantee the P H is equal to the peak. A. So those are our two tasks. The information that we're starting with our buffer solution is H L C L. With 2.20 polarity and Armel Arat e of r K L C L is 0.90 more. I forgot my mole arat e on my acid. So first, the pH and the pH will be equal to the K A. We looked up So the negative log of 3.3 or 3.5 times 10 to the minus eighth and my ratio will be the log of 0.90 over 0.20 Easy enough to solve. The pH is 8.11 8.11 Now, in order for the pH to be equal to the peak a. The ratio of the log of the ratio base over acid has to be equal to one because the log of one is zero. There, I'm writing. The log of base over acid has to equal zero. So the base concentration must equal the acid concentration, which they do not right now. But what can we do to get them thio equal one or equal each other? I should say, Well, we're going to have to add acid because we need to add the accident concentration increase on the base concentration to decrease. It looks like we're gonna have to add 0.35 moles of HCL. And that looks like this problem is solved once I get that colored in.

First let us Cal Khalid B K A. For this value. So the foreigner we are using its peak a is equal to minus off. Log off. Hey, so the final be care value is seven point for six. Ex used the Henderson aggression to solve the it This is the hand disintegration immigration. After substituting the values in the creation, we get a B it off eight point cool. Then he h is equal to the value off B A. What offer? Transit fiction off acid is equal to the concentration off this here. The acid concentration needs to be increased by the base. Concentration needs to be decreased in order to get both have equal concentration. What's your problem? More asset is needed to be added in the buffer solution for the last, but even in the table, our initial concentration Auntie changing concentration after adding the acidic US and at P h equals toe peek A. Both asset and bass have same concentration. Hence we can solve that and get value off X, which equals Toby. Zero point 35 m. Therefore, 0.35 mole off asset. It still must be added to initial buffer. The concentration or a said the concentration off acid on base eventually equal to one another.

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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