If a 1.0 mm of 10.0 NaOH normal is added to a liter of buffer repair in the NaH2O4 Which is 0.042 and Na2HPO4 which is 0.058, how much will PH change

Explanation:

Using H-H equation for the NaH2O4/Na2HPO4 buffer:

pH = pKa + log [Na2HPO4] / [NaH2PO4]

Where pKa of this buffer is 7.198, [Na2HPO4] = 0.058M = moles [NaH2PO4] = 0.042M = moles

Replacing:

pH = 7.198 + log [0.058M] / [0.042M]

Initial pH = 7.34

Now, NaOH reacts with NaH2PO4 as follows:

NaOH + NaH2PO4 → Na2HPO4 + H2O

That means, the moles of NaOH are additional moles of Na2HPO4 and subtracted moles of NaH2PO4.

The moles of NaOH added:

1.0mm³ = 1.0x10⁻³L * (10mol / L) = 0.01 moles

NaH2PO4:

0.042mol - 0.010mol = 0.032moles

Na2HPO4:

0.058mol + 0.010mol = 0.068moles

Replacing in H-H equation:

pH = 7.198 + log [0.068M] / [0.032M]

Final pH = 7.53

Change in pH = 7.53 - 7.34

The change in pH is 0.19 units

it is b that is the answer.