Chemistry Corner

Web Search:

	Groups
			Free Forum Hosting

Important Announcement The MSN Groups service will close in February 2009. You can move your group to Multiply, MSN’s partner for online groups. Learn More
Chemistry Corner	[email protected]

What's New

Welcome Page

About This Site

Message Boards

General

Inorganic

Organic

Pictures

Random

FOR ALL

Handy Symbols

Chemistry Humor

Documents

Organic Sites I

Online Problems

Equations I

Equations II

Eq. Exercises I

Eq. Exercises II

The Mole I

The Mole II

Mole Exercises

Stoichiometry

Stoich. Exercises

More Communities

School's Out!

_________________

Site Map

Tools

General : A few Questions
Choose another message board

Message 1 of 4 in Discussion

From:

N_2006 (Original Message)

Sent: 3/19/2008 4:04 AM

Hey Steve, I have a few questions that I have no idea how to solve!!

First:

As a technician in a large pharmaceutical research firm, you need to produce 250

of 1 $M$ potassium phosphate solution of pH = 7.31. The ${\rm p}K_{\rm a}$ of $\rm H_2PO_4^{~-}$ is 7.21. You have 2 $\rm L$ of 1 $\rm M~KH_2PO_4$ solution and 1.5 $\rm L$ of 1 $\rm M~K_2HPO_4$ solution, as well as a carboy of pure distilled $\rm H_2O$ . How much 1 $\rm M~KH_2PO_4$ will you need to make this solution?
I have tried this over and over.. and I keep getting the wrong answer!!!

Another quesiton:
A certain weak acid, $\rm HA$ , with a

of $5.61 \times 10^{-6}$ , is titrated with $\rm NaOH$ .A solution is made by mixing 9.00 millimoles of $\rm HA$ and 3.00 millimoles of base. What is the resulting pH?

Some how I got 0.48 as the pH.. which I didn't think was right, and it turns out it isn't

I took the Ka and converted to pKa which is 5.25. But everytime I take the concnetrations and do the ICE chart it never comes up with the right answer.

First Previous 2-4 of 4 Next Last

Message 2 of 4 in Discussion

From:

·Steve·

Sent: 3/19/2008 4:59 AM

First Question

This is going to be one of those two equations, two unknowns type problem. First, plug into the Henderson-Hasselbalch equation to calculate the ratio [K2HPO4] / [KH2PO4] :

7.31 = 7.21 + log { [K2HPO4] / [KH2PO4] }

[K2HPO4] = 1.2589
[KH2PO4]

Since both are in the same solution with one total volume, we can simply use moles of substances in the last term of the Henderson-Hasselbalch equation. So,

moles of K2HPO4 = 1.2589
moles of KH2PO4

Or, rearranging,

moles of K2HPO4 = 1.2589 X moles of KH2PO4

Molarity is volume X molarity. Letting "V_K2" be the volume of K2HPO4 and "V_K" be the volume of KH2PO4, and "M_K2" and "M_K" are the corresponding molarities, we have

V_K2 M_K2 = 1.2589 V_K M_K

But, the molarities of the two solutions are the same, 1 M, so M_K2 = M_K. Therefore,

V_K2 = 1.2589 V_K Equation 1

Another note: Since each solution is 1 M, and the desired buffer solution is also supposed to be 1 M in total phosphate concentration, we do not have to make any dilutions. That is, no matter what amount of each solution we mix together, the total phosphate concentration will still be 1 M after mixing.

OK, for the second equation, we know that the total volume is 250 mL. So....

V_K + V_K2 = 250 Equation 2

From Equation 2,

V_K2 = 250 �?nbsp; V_K

Substituting into Equation 1, we have

250 �?nbsp; V_K = 1.2589 V_K

Solve for V_K. V_K2 will be 250 �?V_K, and there will be your two volumes!

I'll look at the second question in the next post.

Steve

Message 3 of 4 in Discussion

From:

·Steve·

Sent: 3/19/2008 5:23 AM

Second Question

First thing to do is calculate the amounts of substances after the reaction with NaOH :

HA (aq) + NaOH (aq) ––�?gt; NaA (aq) + H₂O (l)

Initially: 9.00 mmol 3.00 mmol 0 (solvent)
excess limiting

After Rxn: 6.00 mmol 0 3.00 mmol (solvent)

So, the resulting solution contains 6.00 mmol of HA and 3.00 mmol of NaA. This is a weak acid HA plus its conjugate base A^{�?/SUP> (Na⁺ is a spectator ion here), which is a classic buffer solution.}

Buffer solution = we can use Henderson-Hasselbalch to calculate the pH.

pH = pK_a + log { [A^{�?/SUP>] / [HA] }}

In the log term on the right side, we can use moles of A^{�?/SUP> and HA instead of molarities. Each molarity would be the moles divided by the total volume, but the numerator and denominator would be divided by the same total volume. Therefore, volume cancels and [A^{�?/SUP>] / [HA] = moles of A^{�?/SUP> / moles of HA. Or we can use millimoles; the ratio will still have the same value. Convenient, but we have to be careful, because we usually want to use molar concentrations when working pH problems. But because this is a ratio of molarities in the same solution, the volume cancels and we can just use moles of substances in this case.}}}

So, plugging in, we have

pH = –log (5.61 X 10^�?) + log (3.00 mmol / 6.00 mmol)

from which you can get the pH.

Steve

Message 4 of 4 in Discussion

From:

N_2006

Sent: 4/9/2008 3:07 PM

thank you steve.. again very helpful

First Previous 2-4 of 4 Next Last

Return to General