You are taking samples at a contaminated mine site from a carbonate aquifer to d

Question

You are taking samples at a contaminated mine site from a carbonate aquifer to determine the concentration of dissolved lead in the sample. To preserve the sample, you must acidity it to pH 5. How much HCL, in equivalents/L, must be added to this sample to make it pH 5 if the total carbonate concentration is 10^-2 M and initial pH is 10? What are the acid-base pairs responsible for buffering in this sample You are taking samples at a contaminated mine site from a carbonate aquifer to determine the concentration of dissolved lead in the sample. To preserve the sample, you must acidity it to pH 5. How much HCL, in equivalents/L, must be added to this sample to make it pH 5 if the total carbonate concentration is 10^-2 M and initial pH is 10? What are the acid-base pairs responsible for buffering in this sample You are taking samples at a contaminated mine site from a carbonate aquifer to determine the concentration of dissolved lead in the sample. To preserve the sample, you must acidity it to pH 5. How much HCL, in equivalents/L, must be added to this sample to make it pH 5 if the total carbonate concentration is 10^-2 M and initial pH is 10? What are the acid-base pairs responsible for buffering in this sample

Explanation / Answer

Initial pH=-log(H3O+)=10

[H3O+]=10^-pH=10^-10= 10^-10 moles/L

If pH=5 then [H3O+]=10^-5=10^-5 moles/L

The [H3O+] has to be increased by 10^-5 moles/L

At pH=10 , CO32- +H3O+HCO3- +H2O ,pka=10.35

The solution has buffer acid-base pair CO32- /HCO3-.

pH=pKa + log[CO3 2-]/[HCO3-]

10=10.3+log [CO3 2-]/[HCO3-]

0.3=log[CO3 2-]/[HCO3-]

10^0.3=[CO3 2-]/[HCO3-]

2=[CO3 2-]/[HCO3-]

Or, [CO3 2-]=2*[HCO3-]

So [HCO3-]=10^-2/2= 0.001/2=0.005 M

At pH=5

CO32- +H+HCO3- pka=6.35

5=6.35 +log[HCO3-]/[H2CO3]

-1.35= log[HCO3-]/[H2CO3}

[HCO3-]/[H2CO3}=0.045

[H2CO3]=[Hcl] added as it is the fully neutralized ion at pH=5

[H2CO3}=[HCl]=0.005/0.045=0.1M =0.1 *equivalent/L

(molarity of HCl=Normality of Hcl]

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