hmmm this is not making sense at the moment.
the question:
How many milliliters of 0.246 M HNO3 should be added to 213 mL of 0.006 66 M 2,2-bipyridine to give a pH of 4.19?
the equation:
HNO3- + 22bipy --> Protenated bipy + NO3 2-
u want to find the moles of HNO3...
there are 1.1419 mmoles of bipy...
tried doing an ice table,
H A- HA
I 1.1419 1.1419 0
C -x -x +x
E 1.1419-x 1.1419-x x
and then doing the henderson hasselbach eq but it didn't work out...
pkA of bipy is 4.34.
pH = pka + log ([moles A-] / [moles HA])
4.19=4.34 + log [1.1419-x]/x
nevermind, all clear on this matter, i guess i should try harder before posting on here haha
To solve this problem, we can use the Henderson-Hasselbalch equation to relate the pH of the solution to the ratio of the concentrations of the conjugate base (A-) and the weak acid (HA). Given that the pKa of bipyridine (HA) is 4.34, we can rewrite the Henderson-Hasselbalch equation as:
pH = pKa + log([A-] / [HA])
Now, let's plug in the values we have:
pH = 4.34 + log([1.1419 - x] / x)
We want to find the value of x, which represents the moles of HNO3 added. However, in order to solve this equation, we need to make some simplifications.
First, we need to convert the concentrations of HNO3 and 2,2-bipyridine to moles. We have 213 mL of 0.00666 M 2,2-bipyridine, so the moles of 2,2-bipyridine can be calculated as follows:
moles of 2,2-bipyridine = concentration * volume
= 0.00666 M * 0.213 L
= 0.00142 moles
Next, we can solve for the moles of the conjugate base (A-) and the weak acid (HA) using the ice table approach. We have already set up the initial, change, and equilibrium rows of the table:
H A- HA
I 1.1419 1.1419 0
C -x -x +x
E 1.1419-x 1.1419-x x
From the equation, we can see that moles of A- and HA are the same, which is represented as (1.1419 - x). The change in the moles of HA is +x, which represents the moles of HNO3 added.
Now, let's proceed with the calculations. We need to equate the expression [A-] / [HA] to the ratio of (1.1419 - x) / x and then solve for x:
4.19 = 4.34 + log[(1.1419 - x) / x]
To isolate the x term, we subtract 4.34 from both sides and rewrite the equation:
log[(1.1419 - x) / x] = -0.15
Next, convert the equation from logarithmic form to exponential form:
(1.1419 - x) / x = 10^(-0.15)
Simplify the right side:
(1.1419 - x) / x = 0.707
Now, cross-multiply:
1.1419 - x = 0.707x
Rearrange the equation:
1.1419 = 1.707x
Divide both sides by 1.707:
x ≈ 0.6696
Therefore, approximately 0.6696 moles (or 669.6 millimoles) of HNO3 should be added.
I hope this explanation helps you understand the process of solving this problem! Let me know if you have any further questions.