Acetic Acid Lab Report

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BIOC 2200 – Experiment 1: Buffers and the effect on amino acids

Aiden Forsyth

1009696399

Wednesday PM

Bench 13

Results

Questions
A1.
By the Henderson-Hasselbalch equation (Plumber, 2004).

pH=pka+log⁡〖([HA])/([A^-])〗 n_([HA])=M*V=0.10 mol/L*0.050L=0.0050 mol n_[A^- ] =n_NaOH=M*V=0.20 mol/L*0.001L=0.0002 mol

Since acetic acid is transformed to sodium acetate there is a consumption of [HA] by the amount of moles of sodium hydroxide. pH=4.76+log⁡〖0.0002/(0.0050-0.0002)〗=4.76-1.38=3.38 A3.

To completely dissociate all of the acetic acid with sodium hydroxide to sodium acetate there must be zero moles of acetic acid remaining (Plumber, 2004).

From A1 we know that the number of moles of acetic acid is
While the pka for acetic acid can be determined in Graph 1 to be 4.73 the volume of sodium hydroxide in 3mL short of the calculated 13mL point. The published value of 4.76 is similar to the observed pka of acetic acid (Horton et al., 2006). The pka of boric acid was observed to be 9.27 at 14mL of sodium hydroxide added which is consistent with the published value of 9.27 (Silberberg, 2010). Boric acid would be a good buffer for an experiment conducted at a pH of 8.5 because the mixed solution would have a similar concentration of acid and conjugate base. Thereby resisting a change to its pka or 9.27. Acetic acid would have very little resistance to a change in the experiments pH because its pka is 4.76 thus being an already defeated buffer by the time we start the experiment at a pH of
To fully dissociate an amino acid is to ionize all of the present ionizable groups (Horton et al., 2006). To such an end the environment’s pH must be higher than the highest pKa so as to deprotonate the last remaining group. The pKas for arginine are 1.8 for the alpha carboxyl group, 9.0 for the alpha amino group and 12.5 for the R group. So the aim is to have an environment with a pH higher than 12.5 to fully dissociate arginine. The corresponding volume of NaOH require to have a pH of 12.52 was 79mL. There must be three equivalents of NaOH to dissociate all three of the groups found in arginine where as in acetic and boric acid there are only two groups and therefore only need two NaOH equivalents. Valine is similar where it has two groups that can be ionized with pKas of 2.3 for the alpha carboxyl group and 9.7 for the alpha amino group (Plumber, 2004). For 1 mol of valine the required the volume of 0.20M NaOH needed is

pH=pKa+log (〖[A〗^-])/[HA]
〖[A〗^-]=〖10〗^((9.8-9.7) )*1=1.3mol so 1.3 mol of NaOH is required

D3.

The two forms of threonine found in milk at a pH of 6.5 would exist in a ratio based on their isoelectric point (Hine et Martin, 2014). pI=(∑▒pKa)/2=(2.1+9.1)/2=5.6 Because the environment’s pH is less than the isoelectric point the two forms of threonine present are the zwitterion and cation forms. The ratio of the two being pH=〖pKa〗_cation+log (〖[A〗^-])/[HA]

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