Determine the % of Al3+ in antacid sample (0.6495 g) 1) Assay for Al3+ in antaci

Question

Determine the % of Al3+ in antacid sample (0.6495 g) 1) Assay for Al3+ in antacid sample

Average of titre volume = 19.50 Blank 24.80 ml

Concentration of excess EDTA in 25 mL = (0.009800 mol/L x 0.025 L) /1.00 L = 0.000245 mol/L

Concentration of ZnSO4 in 19.50 mL = (0.0100 mol/L x 0.0195 L) / 1.00 L = 0.000195 mol/L

Blank????

Number of mol of DETA that react with Al3+ = ???

the concentration of Al 3+ in the sample of antacid (0.6495 g) =???mol L-1

M.w of Al3+ = 26.98153 g/ mol

Mass Of AL3+ = ????

% = ???? x 0.6495 g /100 = ?

Note: it should be around 6.70%

Explanation / Answer

Determine the % of Al3+ in antacid sample (0.6495 g) 1) Assay for Al3+ in antacid sample

Average of titre volume = 19.50 Blank 24.80 ml

Concentration of excess EDTA in 25 mL = (0.009800 mol/L x 0.025 L) /1.00 L = 0.000245 mol/L

Concentration of ZnSO4 in 19.50 mL = (0.0100 mol/L x 0.0195 L) / 1.00 L = 0.000195 mol/L

Number of mol of DETA that react with Al3+ 0.00161

the concentration of Al 3+ in the sample of antacid (0.6495 g) =0.00161 mol L-1

M.w of Al3+ = 26.98153 g/ mol

Mass Of AL3+ =0.0433

% = 0.0433/0.6495 g *100 = 6.68 % Answer

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