buffer assignment 1.png - Buffer Writing Assignment A pH 4.30 acetate buffer(pKa is 4.74 has a total concentration of 0.200 M(the concentration of

buffer assignment 1.png - Buffer Writing Assignment A pH...

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Unformatted text preview: Buffer Writing Assignment A pH 4.30 acetate buffer (pKa is 4.74) has a total concentration of 0.200 M (the concentration of acetic acid plus acetate is 0.200 M) and a volume of 100.00 mL. 1. How much 0.100 M NaOH must be added to reach the equivalence point? We know that the ph of the buffer is equal to 4.30, and we know that the pka of acetic acid is equal to 4.74. We can also determine the total moles of the buffer by multiplying the molarity of the buffer with the volume of the buffer (after converting from ml to L): 0.200 M * 0.100 L = 0.02 moles buffer With this information, the Henderson Hasselback equation can be used to determine the moles of acetic acid and acetate. In the following equation "*" refers to the moles of acetate. PH = pK. + log (moles weak base / moles weak acid) PH = pKa + log (moles acetate / moles acetic acid) 4.30 = 4.75 + log(x/[0.02 - x)) 104 30-4.74 = x/(0.02 - x) 0.36307805477 = x/(0.02 -x) 0.00726156109 - 0.36307805477 * x =x 1.3630780547 * x = 0.00726156109 * = 0.00532732594 moles acetate Thus, we have 0.0053273 moles of acetate. The moles of acetic acid can be found by subtracting the moles of acetate from the total moles of buffer: Moles of acetic acid = 0.02 - 0.00532732594 Moles of acetic acid = 0.01467267406 At the equivalence point, we know that the moles acetic acid is equal to the moles of strong base that is added to the buffer. We know the molarity of the strong base (NaOH, in this case) is 0.100M....
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  • Winter '07
  • Pang
  • pH, total moles of the buffer, 0.0053273 moles

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