Calculate the change in ph if 0.070 g of solid naoh is added to 100 ml of the solution in the problem above.

Assuming the pH of a buffer solution that contains 0.65 M NaH2PO4 and 0.11M Na2HPO4 is 6.44.

Calculate the change in pH if 0.070 g of solid NaOH is added to 100 mL of the solution in the problem above

∆ pH = 0.08

Explanation:

Calculate concentration in mmoles of each species before adding the NaOH.

[NaH2PO4] = M NaH2PO4 x mL NaH2PO4 = (0.65)(100) = 65 mmoles NaH2PO4

[Na2HPO4] = M Na2HPO4 x mL Na2HPO4 = (0.11)(100) = 11 mmoles Na2HPO4

When you add the strong base NaOH to the buffer, it will react with the acid part of the buffer (i.e. the species with the most Hs --- NaH2PO4).

0.070 g NaOH x (1000 mg / g) x (1 mmole NaOH / 40.00 mg NaOH) = 1.75 mmoles NaOH

[Concentration] . . . . . .NaH2PO4 + NaOH ➡ Na2HPO4 + H2O

initial . . . . . . . . . . 65 . . . . . .1.75 . . . . . . . . .11

change . . . . . . . .-1.75 . . . . .-1.75 . . . . . . .+1.25

final . . . . . . . . . . .63 . . . . . . . .0 . . . . . . . . . .13

pKa for H2PO4- = 6.2 x 10^-8; pKa = 7.21

pH = pKa + log ( [Na2HPO4] /[NaH2PO4]) = 7.21 + log (13/ 63) = 7.21 - 0.69 = 6.52

∆pH is minute (from 6.44 to 6.52). = 0.08

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.