How to find the pH to...

Question

1.Find the pH of a buffer that is 0.150 M lactic acid (HC3H5O3, “HLc”) and 0.130 M sodium lactate

(NaC3H5O3, “NaLc”). Ka for HLc = 1.4×10-4.

2.Find the pH of a 0.10 M OCl- solution. Ka of HOCl is 3.0×10-8.

Our teacher gave us the answers to the questions but we can't seem to find out how to get to the answer. What are the steps to finding the answer? help please! thanks

Answer 1

1. Calculate the pH of a buffer that is 0.12 M in lactic acid (Ka = 1.4 x 10-4) and 0.11 M in sodium lactate.  ans. 3.82



  2. A solution is made by mixing 30. mL of 0.15 M CH3COOH (Ka =1.8 x 10-5) with 70. mL of 0.20 M CH3COONa.  Calculate the pH of this solution.  ans. 5.24



  3. Calculate the pH of a buffer formed by mixing 85. mL of 0.13 M lactic acid (Ka = 1.4 x 10-4) with 95 mL of 0.15 M sodium lactate.   ans. 3.96



  4. 20.0 g of CH3COOH (Ka = 1.8 x 10-5) and 20.0 g of CH3COONa are dissolved in 2.00 L of solution.  Find the pH.  What affect does adding water have on the pH?  ans. 4.60, no effect.

Answer 2

begin by writing out the equation for the dissociation of lactic acid into the H+ and the lactate. The Ka value tells you how much of the lactic acid will dissociate. If it is a strong acid, the Ka will be very high b/c most of the sol'n will be acid. A weak acid will have a small Ka b/c most of it remains in the undissociated form so there will be very little free [H+]

[lactic acid] = Ka X  [Na-lactate] X [H+]

Now you plug in the numbers that you are given into the equation and solve mathmatically for the unknown [H+]

0.150 M [Lactic acid] = 1.4 E-04 X (0.130 M) [Lactate] X [H+]

[H+] = 1.4 E-04 X (0.130 M) / (0.150 M) = ???? need calculator

Now that you have the [H+] you can apply the formula for pH which is the pH = - log [H+] = ????  I don't have a calculator

For question 2 you are going to do the exact same thing.

Always begin by writing out the equation and you will never go wrong.

They rewrote the question to trick you by writing certain things backwards. Always begin with the acid and dissociating into the base pair and the [H+].

[ClOH] = Ka X [ClO-] X [H+]

Now substitute in all the numbers that you are given.

Not so easy is it? Let's remember the principle. How do you get the value of Ka? It is the ratio of the [Acid] / [ Base complement]. They wrote this question to trick you into remembering some basic concepts of chemistry.

so now you have the new equation

[ClOH] / (0.10 M) [ClO-] = Ka = 3.0 E-08

So now we solve the equation to find out how much [ClOH] we have

[ClOH] = 0.10 M [ClO-] x 3.0 E-08 =  ???? need calculator

When you plug this into the calculator, you now have the value of [ClOH] and put that value into the original dissociation equation and solve for [H+]

[H+] = Ka x (0.10 M) [ClO-] / ???? [ClOH] which we found from above

Now you have the [H+] so you can use the pH = -log [H+]

I don't have a calculator so you'll have to plug the numbers in.

Easy, eh ?