BuffersPostLab.docx - Buffers Post Lab Calculations 1 2−...

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Buffers Post Lab Calculations 1) 2 ¿ H PO 4 ¿ ¿ ¿ H 2 PO 4 ¿ ¿ ¿ ¿ pH = pK a + log ¿ First Buffer 2 ¿ 0.20 M H PO 4 ¿ ( 0.0208 L ) ¿ 0.20 M H 2 PO 4 ¿ ( 0.0297 L ) ¿ ¿ ¿ ¿ ¿ pH = 7.21 + log ¿ Initial pH of firstbuffer = 7.06 Second Buffer 2 ¿ 0.20 M H PO 4 ¿ ( 0.0303 L ) ¿ 0.20 M H 2 PO 4 ¿ ( 0.0205 L ) ¿ ¿ ¿ ¿ ¿ pH = 7.21 + log ¿ Initial pH of second buffer = 7.38 2) Buffer Capacity = ∆ N ∆ pH
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First Buffer β = | ( ( 0.00497 LHCl )( 0.15 M HCl ) .1 Lbuffer solution ) 5.99 6.69 | = 0.01065 mol L∙ pH Buffer capacity of first buffer after first additionof acid = 0.01065 mol L∙ pH 3) Buffer Capacity = ∆ N ∆ pH Second Buffer β = | ( ( 0.00499 L HCl )( 0.15 M HCl ) .1 Lbuffer solution ) 6.43 6.93 | = 0.01497 mol L∙ pH Buffer capacity of second buffer after first additionof acid = 0.01497 mol L∙ pH 4) Buffer Capacity = ∆ N ∆ pH First Buffer β = | ( ( 0.00518 L NaOH )( 0.15 M NaOH ) .1 Lbuffer solution ) 7.16 6.62 | = 0.0144 mol L∙ pH Buffer capacity of first buffer after first additionof base = 0.0144 mol L∙ pH 5) Buffer Capacity = ∆ N ∆ pH
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Second Buffer β = | ( ( 0.00503 L NaOH )( 0.15 M NaOH ) .1 Lbuffer solution ) 7.95 6.91 | = 0.00725 mol L∙ pH Buffer capacity of second buffer after first additionof base = 0.00725 mol L∙ pH 1) The purpose of the experiment was to determine the buffer capacity of two different buffer solutions of different initial concentration ratios of ¿ H 2 PO 4 ¿ and 2 ¿ H PO 4
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  • Spring '08
  • LEMASTER
  • pH, buffer solution, First Buffer, Initial pH of first buffer

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