Question: Base: NaH 2 PO 4 * H 2 O = 137.99 g/mol Acid: Na 2 HPO 4 * 7H 2 O = 268.07 g/mol pH= 8.00 Conc (M) =0.100 8.21+Log[A/B] 0.79=Log[A/B] -> 10...
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# Question:Can you please help me with part a, b, & c based on the information below:242Acid: Na24 *2pH= 8.00

Base: NaHPO * HO = 137.99 g/mol

HPO 7HO = 268.07 g/mol

Conc (M) =0.100

8.0=7.21+Log[A/B]

0.79=Log[A/B] -> 100.79=B/A -> 6.16595=B/A -> B=6.16595A

0.01mol= A+6.16595

0.01mol= 716595

Acid = 0.1925 g

Base = 2.30 g

Solids (acid & base) were added to a 100ml volumetric flask and filled with DI water to the 100mL mark

a) Calculate the theoretical pH of the buffer solution after the addition of 1.00mL of 0.100M NaOH ;added to 100.00ml of buffer solution

b) Calculate the theoretical pH of the buffer solution after the addition of 1.00mL of 0.100M HCl

c) Calculate the theoretical values for the pH of the NaOH and HCl solutions made in water

-1.00mL of 0.100M NaOH added to 50.0mL of DI water - pH: 11.11

-1.00ml of 0.100M of HCl to 50mL of DI water - pH: 2.56

i. You will need to know the concentration of NaOH an HCl once they were added to the water in each solution.

ii. You will then need to know that HCl and NaOH are strong acid and base, respectively, which means they will dissociate completely in water

iii. You can then determine the [H+] concentration (or[OH-]... then the [H+]) and calculate pH