Using equations showing the relationship between [acid], [conjugate base], and [

Question

Using equations showing the relationship between [acid], [conjugate base], and [buffer concentration], mathematically show the negligible change in pH when 2 microliters of 10M HCl is added to 100 mL of 50 mM phosphate buffer at pH 6.9. To address this problem, the following steps will be needed: (i) You will have to calculate the pH of the buffer after adding the 2 microliters of 10 M HCl (ii) To do this, calculate the new concentration of [acid] and [conjugate base] after the addition of 2 microliters of 10 M HCl, and (iii) To appreciate the effect of buffers, show the pH change in 100 mL of water after adding 2 microliters of 10 M HCl that does not contain the phosphate buffer ions. (show all calculation steps)

Explanation / Answer

(i) calculate the pH of the buffer after adding the 2 microliters of 10 M HCl

mol of HCl added=2*10^-6 L*10M=2*10^-5 mol

total mol of phosphate buffer=0.1L*50*10^-3 mol/L=5*10^-3 mol

Using henderson hasselbach equation,,

pka of NaH2PO4/Na2HPO4=7.21 [best phosphate buffer for pH range 6.0 to 8.0]

pH=pka+log[base]/[acid]

6.9=7.2+log[base]/[acid]

log[base]/[acid]=-0.3

[base]/[acid]=0.5

But [base]+[acid]=5*10^-3 mol

So,[base]==[Na2HPO4]=(0.5/1+0.5)*5*10^-3 mol=1.7*10^-3 mol

[acid]=[NaH2PO4]=(1/1.5)**5*10^-3 mol=3.3*10^-3 mol

mol of base neutralized by HCl=2*10^-5 M [so ,more acid NaH2PO4 is formed by neutralization]

Na2HPO4 +HCl --->NaH2PO4 +NaCl

mol of acid remaining=3.3*10^-3 mol+2*10^-5 M=3.3*10^-3 mol+0.02*10^-3 M=3.32*10^-3 mol

mol of base remaining=1.7*10^-3 mol-0.02*10^-3 M=1.68*10^-3 mol

new pH=pka+log [base]/[acid]=7.2+log (1.68*10^-3 mol/3.32*10^-3 mol)=6.9

pH =6.9(same as before)

ii)new concentration of [acid] and [conjugate base] after the addition of 2 microliters of 10 M HCl,

[NaH2PO4]=3.32*10^-3 mol/total volume=3.32*10^-3 mol/0.1L=3.32*10^-2 mol/L

[Na2HPO4]=1.68*10^-3 mol/0.1L=1.68*10^-2mol/L

iii) pH of water=7.0

pka of water=15.7

ka=kw=[OH-][H3O+]=ionic product of water=10^-14

On addition of HCl,

mol of HCl added=2*10^-6 L*10M=2*10^-5 mol=mol of H3O+ added

[H3O+]=2*10^-5 mol/0.1L=2*10^-4 mol/L

[OH-]=1*10^-7 /L (neutral wter)

new [H3O+]=2*10^-4 +1*10^-7mol/L=2*10^-4 mol/L

new[OH-]=1*10^-7M (no change as no neutralization takes place)

kw=[H3O+][OH-]

10^-14 /[H3O+]=[OH-]=10^-14/2*10^-4=0.5*10^-10

pOH=-log[OH-]=

pH =14-pOH=14-10.3=3.7

pH has changed a lot from 7.0 to 3.7 for water after adding 2 microliters of 10 M HCl

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