please answer the follwing questions 1-A 200mL sample of water required 25.12 mL

Question

please answer the follwing questions

1-A 200mL sample of water required 25.12 mL of a 0.0200 N standard H2SO4 for titration to the methyl orange endpoint, pH 4.5. What was the total alkalinity of the original sample?

2-A water sample was run through the colorimetric procedure for the analysis of nitrate, giving 55.0% transmittance. A sample containing 1.00 ppm nitrate run through the exactly identical procedure gave 24.6% transmittance. What was the concentration of nitrate in the first sample?

3-Analysis of a leadcontaining sample by graphitefurnace atomic absorption analysis gave a peak of 0.075 absorbance units when 50uL of pure sample was injected. Lead was added to the sample such that the added concentration of lead was 6.0ug/L. Injection of 50uL of ‘spiked’ sample gave an absorbance of 0.115 absorbance units. What was the concentration of lead in the original sample?

Explanation / Answer

At pH =4.5, we get total alkalinity and this is determined by

and the following reaction occurs

HCO3- + H+ H2CO3

Total Alkalinity

b) %T= e(-ebc)

e= extinction coefficien, b= path lenght and c= concentration

lnT1 = -ebc1

lnT2 = -ebC2

lnT1/lnT2 = C1/C

ln(55)/ln (24.6)= C1/1

C1= 1.25 ppm

c) A= ebC

0.075= eb* C1

0.115 =ebC2

ebC2/ebC1= 0.115/0.075

C2/C1=1.53

C2= 6*1.53= 9 ug/L

= Amount of acid used to reach pH 4.5 (ml) * Normality of acid (eq/L) * 50,000 (mg CaCO3/eq) / sample volume (ml)

Total alkalinity= 25.12*0.04 (Eq/l)*50000 mgCaCO3/eq)/200 =251.2 mg/L

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