What is the pH of a buffer solution prepared by dissolving 0.250 mol of NaH2PO4 and 0.075 mol of NaOH in enough water to make 1.00 L of solution? (Ka (H2PO4-) = 6.2 X 10^-8)
(a) 6.32
(b) 6.83
(c) 7.21
(d) 7.71

Question

What is the pH of a buffer solution prepared by dissolving 0.250 mol of NaH2PO4 and 0.075 mol of NaOH in enough water to make 1.00 L of solution? (Ka (H2PO4-) = 6.2 X 10^-8) 
(a) 6.32
(b) 6.83
(c) 7.21
(d) 7.71

Accepted Answer

Hello. Everyone send this video. We have 1.5 liters of buffer solution that was made by adding this many moles of r K O H to this many moles of R K H two C six H 507. Here in this question was being asked what is not or what is the ph of the buffer solution? And were given the K value of the H two C six H minus with this value here. Alright, so first of course we need to write out the reaction. So starting off with my star materials, this is K H two C six H507. And this will go ahead and break down into our K plus and H two C six Age 507 -. Alright, so K plus is a main group element. There is a charge less than plus three which is neutral. So Okay. Plus this is neutral as for the H two C six H 507 minus. This can act as a weak base or a weak acid. So just put W B for weak base or can be a weak acid. So W. A. All right. So for our K O H. This breaks down into K plus and O H minus. He already said that K plus is a main group element but they charge less than plus three which is neutral. As for R O H. Here this is the strongest base in a grid solutions. So I just write strong base and K plus will not be part of the reaction. Therefore let's do it like this here. Our reaction then is going to be a church to C six H 507 minus and sis acquis, it reacts with our base here. Oh H minus which of course is our equilibrium areas. And from the product side we have H. C six H 507 minus As well as our liquid water. So H20 liquid. Alright, so now let's determine the equilibrium concentration. We're gonna go ahead and construct a ice table. Alright, so again let's rewrite the equation. We have H two C six H 507 minus again, this is a quiz. It reacts with our base O. H minus, drawing the equilibrium arrows point product side. Then we have hc six H 507 minus minus which is a quiz. And then the liquid water of H 20. So we have I. For initial C. For change and E. For equilibrium. Alright, so for my first story material We have zero points to 75 moles. Then for my O. H. - Here this is 0.085 moles. Initially for my product side we have zero and since water is a liquid we ignored for the rest of our ice table. All right now for the change we subtract. So we're subtracting 0.085 moles. Same thing for my OH -0.085 moles. And then for my products I we're adding, so we're adding 0.08 five moles. Alright then from equilibrium is basically just combining this and this and this. So if we're taking 0.275 - in 0.085 At equilibrium we get 0.190 moles from my first product then from OH - is of course zero. And then from the product side we have 0.085 moles. So our H two C six H 507 minus. This is a weak acid. Maybe we can actually write this out. So H two C 6 H -. This is a weak acid. As for my H. C. Six H 5072 minus, this is one less H plus than the weak asset that we have. So this is a conjugate base of H. Two C six H 507 minus. Alright so since we are dealing with a weak base here we can deal or we can use the Henderson Hasselbach equation to discover our ph so let's actually write out this equation. So we have P. H. Equaling to the P. K. A. Plus the log of the concentration of our conjugate base over the weak acid. Alright, so first of all let's go ahead and realize what R. P. K. Is. We are given the K. Value. So we just need to find the P. K. So for P K. A. Is equal to the negative log of R. K. A value which is already given. So we just need to put it into this equation. The K value that's given is 1.7 times 10 to the negative five. So once you put that into a calculator we will get the peak a equaling 24.76955. As for the concentration of our contra Git base. This is of course just h. c. six H 5072 minus. So this is just 0.85 moles which we got from the ice table. We need to go ahead and divide this with the volume and that is 1.50 liters. And that's how we get concentration since concentrations unit is more per liter. So putting that into vocabulary, we get that the concentration of our conjugate base is 0.056667. Now, for the concentration of our weak acid this of course is the concentration of the each. To see let's actually write this a little bit better. All right. C six h 507 - for the moles, we have 0.190 moles for the leaders or the volume here, that's 1.50 L. Once you put that into a calculator to get the numerical value to be 0.12667. Now we go ahead and utilize this Henderson Hasselbach equation. So the P H. Let's see. We're using equation in purple. So let's draw the arrow like. So. All right, so the ph then is equal to basically. Now at this point is plugging all these values that we have in blue. So ph is equal to the P. K. 4. plus the log of. Let's see. So for our contract base this is 0.05667. The concentration of our weak acid is 0.12667. So once you put that into a calculator, we get the value to be 4.4-0-1. Of course, put it into the correct amount of significant figures. We get that the ph here is equal to 4.42. So this right here is going to be my final answer for this problem. Thank you all so much for watching

What is the pH of a buffer solution prepared by dissolving 0.250 mol of NaH2PO4 and 0.075 mol of NaOH in enough water to make 1.00 L of solution? (Ka (H2PO4-) = 6.2 X 10^-8) (a) 6.32 (b) 6.83 (c) 7.21 (d) 7.71

Verified Solution

Key Concepts

Buffer Solutions

Henderson-Hasselbalch Equation

Acid Dissociation Constant (Ka)