How would you make 100mL of tartrate buffer, pH 3.5, from 0.100 M tartaric acid, 1M NaOH and deionized and distilled H2O? You need to consider the ionization of two the two acidic groups separately.
To make 100 mL of tartrate buffer at pH 3.5, we need to consider the ionization of tartaric acid and the neutralization of the resulting acidic groups using NaOH. Here's how you can approach it:
1. Determine the acid dissociation constants (Ka) for tartaric acid. For a diprotic acid like tartaric acid, there are two dissociation constants, Ka1 and Ka2, corresponding to the ionization of the first and second acidic groups, respectively.
2. Calculate the ratios of the acid to base required to achieve the desired pH. For tartaric acid, since there are two acidic groups, both need to be neutralized. The desired pH of 3.5 is closer to the pKa2 value (the negative logarithm of Ka2), so the base required will mainly neutralize the second acidic group.
3. Determine the number of moles of tartaric acid required. This can be calculated using the concentration (0.100 M) and the volume (100 mL).
4. Determine the number of moles of NaOH required. Since we have a 1M solution of NaOH, the number of moles can be calculated using the volume required to neutralize the second acidic group.
5. Calculate the volume of deionized and distilled water required. This is determined by subtracting the volume of tartaric acid and NaOH solutions from the final volume of the buffer (100 mL).
Here's an example of the calculations using arbitrary values:
1. Assume Ka1 = 10^-3, and Ka2 = 10^-5.
2. Calculate the ratios of acid to base for the desired pH of 3.5. Since pH = pKa2 + log(base/acid), rearrange the equation to find the ratio of base/acid: base/acid = 10^(pH - pKa2). In this case, the ratio will be 10^(3.5 - (-5)) = 10^8.5.
3. Calculate the number of moles of tartaric acid: volume (in L) x concentration (in mol/L) = 0.100 M x 0.100 L = 0.010 mol.
4. Calculate the number of moles of NaOH required: ratio of base/acid x moles of acid = 10^8.5 x 0.010 mol = 10.^(8.5 + log(0.010)) = 10534 mol.
5. Calculate the volume of water required: final volume - volume of tartaric acid - volume of NaOH = 100 mL - 0.100 mL - (10534 mol / 1 M) = 100 mL - 0.100 mL - 10534 mL = -10434.100 mL.
Note: The negative volume implies that you'll need to add 10.4341 L (10434.1 mL) of water to reach the final 100 mL volume.
Keep in mind that the specific values (concentration, pKa, etc.) will vary based on the specific system, so use appropriate values for accurate calculations.