Chemistry 5.9 Buffers

Example questions continued.

1000molHF + 0.1molNaF 1000/0.1= 10,000times more acidic. 10^4 3.2-4= -0.8= pH This wouldn't be a buffer because you are outside buffer range (2.2 to 4.2)

Examples Continued

8molHF + 1mol NaF what is the pH 8 times more acid if it was 10 times more acid pH= 3.2-1=2.2 8 is a little smaller than 10 so pH is about 3.2-0.8= 2.4

Buffers

Buffers: A buffer resists changes in pH. -A buffer is composed of a weak acid conjugate base. HF + F- This is a buffer. But there is a problem here because pure F- doesn't exist. You will never get a pure sample of an anion or a cation. You usually add an anion or cation as a salt. -What could you combine with F- that would always be soluble (solubility rules: any group 1 metals, NO3, NH4, ClO4, C2H3O2). -Any group 1 salt is what we would typically use. So you would see it like this: (sodium is just a spectator, ignore it) -You want these to be present in a 1:1 ratio. -A buffer is at its best (maximum buffering capacity) when the weak acid and conjugate base are as high concentrations as possible and as close to 1:1 ratio as possible. -They have to be within 10 fold factor of each other. You could have up to 10 times more of 1 vs the other but thats it. HF + NaF

Ways to make a buffer

HF + NaF -This is what a buffer is but there are a couple of ways you can make it (1) FIRST OPTION HF + any strong base (NaOH) - in this case you have to add them in a 2:1 ratio. NaOH is a strong base so it will dissociate completely in water or react completely with any acids present in the water. HF + NaOH ------> NaF + H2O If you add HF and NaOH in 2:1 ratio, NaOH is the limiting reagent. You would have 1 mole of HF left and 1 mole of NaF formed. So you have 1:1 ratio of HF and NaF. The weak acid (HF) and its conjugate base (F-) are present in a 1:1 ratio which means it is a buffer. (2) Second Option NaF + HCl -------> NaCl = HF Take the weak base (NaF) and add any strong acid (HCl) in a 2:1 ratio. because it is a strong acid, the reaction will go to 100% completion. -Here the limiting reagent is HCl, so you will have 1 mole of NaF left and 1 mole of HF produced. That is a buffer. -When you neutralize half of a weak acid or half of a weak base, then you are half way to the equivalence point in a titration.

How to calculate the pH of a buffer.

You can use the Ka or Kb equation of the buffer. - More commonly for buffer calculations we use: Henderson Haselbach Equation: pH=pKa + log ([A-]/[HA]) -you want ratio of conjugate base (A-) and conjugate acid (HA-) to be 1. -What is the log of 1? Its 0. So if thats all equal to 0, the pH=pKa. Buffers are best when they are in 1:1 ratio and that point also happens to be the point when the pH=pKa Example Question: strong acids: Ka: pKa HF 6.8x10^-4 3.2 HCO2H 1.8x10^-4 3.8 HC2H3O2 1.8x10^-5 4.8 -This tell you that HF mixed with NaF would be at maximum buffering capacity at a pH of 3.2 (when the pH=pKa) -pKa also gives you the range. The range over which a buffer is effective is best at when pka=pH (3.2 in this case) plus or minus 1. -So HF when mixed with NaF (depending on the ratio in which you mix) could be used as buffer anywhere from pH 2.2 to pH 4.2. - So you might be asked which of the following acids, when mixed with its conjugate base could make a buffer of pH 5.1. - You would calculate the pKa for each and see which buffer range pH of 5.1 falls into. -The answer would be HC2H3O2