Question

The virtual lab stockroom contains NaOH, water and an unknown solid weak acid. Perform experiments to identify the pka and molar mass of the bottle of unknown solid acid. You may enter your answer in the form below the Virtual Lab NOTE: You can submit your answers up to three times. If all three answers are incorrect, you will be given the correct answer and asked to reload the page and try a new problem. Good Luck! Please be aware that reloading the page will result in having to start a new problem with a different unknown acid

Data i got:

1g of solid acid=0.51706 ml

50 ml of water used to dissolve

0.5 ml of indicator added to acid solution

8.52 ml of 1M NaOH titrated from burrete into acid solution.

Answer 1

Assuming, the given unknown acid is a monobasic HA.
The neutralization reaction would be:
$NaOH(aq) \, +\, HA(aq)\, \rightarrow \, NaA\, +\, H_{2}O$
Number of moles of NaOH added = Molarity x Vol. in L = 1 x 0.00852 L = 0.00852
Looking at the balanced equation, number of moles of HA added = 0.00852
Let molar mass of HA = M
Using, given mass/Molar mass = number of moles
1/M = 0.00852
M = 1/0.00852 = 117.37 g/mol
The salt formed (NaA) will be a basic salt and the pH of this salt can be calculated by the formula:
$pH\, =\, 7\, +\, \frac{1}{2}\times [pk_{a}\, +\, logc]$
c(concentration of the salt) or [NaA] = (number of moles/vol. of solution) x 1000
Number of moles of the salt formed = 0.00852
PS: It is not clear why vol of the solid acid taken is provided.
vol. of solution =  50 mL of water + 8.52 mL of NaOH + 0.5 mL of indicator = 59.02 mL
(Although the indicator volume is negligible, but since it is provided we can add it)
[NaA] = (0.00852/59.02) x1000 = 0.144
If pH at the end point is known, pKa can be callculated.