Calculate the pH of the solutions obtained by dissolving the following in a liter of water: 0.2 mole of glycine plus 0.1 mole of NaOH
I know the answer is ph = 9.60 which i know is pka2 of glycine but im not sure how to set up this question. This is just a practice problem for my homework which will have similar questions but i would like to know how to do my homework by knowing how to do this question.
Thank you
..NH2CH2COOH + NaOH =>NH2CH2CONa + H2O
I...0.2.........0..........0
add............0.1.............
C...-0.1.......-0.1........+0.1
E....0.1.........0.........0.1
pH = pKa glycine + log (base)/(acid)
base = salt = 0.1
acid = glycine = 0.1
pH = pKa + log 0.1/0.1
0.1/0.1 = 1 and log 1 = 0
so pH = pKa
To calculate the pH of the solution, we need to consider the dissociation of glycine (an amino acid) and the reaction between glycine and sodium hydroxide (NaOH). Here's how you can set up the question:
1. Write the chemical equation for the reaction:
Glycine + NaOH -> Glycine NaOH
2. Calculate the moles of glycine and NaOH used:
Glycine: 0.2 moles
NaOH: 0.1 moles
3. Determine the limiting reagent:
To find the limiting reagent, compare the moles of glycine and NaOH. In this case, NaOH is in excess because we have more moles of glycine.
4. Write the net ionic equation:
Only species involved in the reaction are glycine and water. The net ionic equation looks like this:
Glycine + H2O -> H+ (aq) + OH- (aq)
5. Calculate the initial concentration of glycine and NaOH:
In a liter of water (assuming no volume changes), the concentration of glycine is 0.2 M, and the concentration of NaOH is 0.1 M.
6. Calculate the pH:
The initial concentration of OH- is 0.1 M (as NaOH dissociates completely). Since the OH- concentration is known, we can calculate the pOH using the formula: pOH = -log(OH- concentration).
pOH = -log(0.1) = 1
pH = 14 - pOH = 14 - 1 = 13
Therefore, the pH of the solution obtained by dissolving 0.2 moles of glycine plus 0.1 moles of NaOH in a liter of water is 13.