g 30.0 mL of a buffer is made by mixing 20.0 mL of 0.10 M HNO2 and 10.0 mL of 0.10 M NaOH solution. What is the pH of this buffer after 0.04 mL of 10.0 M hydrochloric acid are added to it

pH = 2.98

Explanation:

pKa of HNO2 is 3.35

HNO2 reacts with NaOH as follows:

HNO2 + NaOH → NaNO2 + H2O

Where moles of NaOH added are equal to NaNO2 produced

And initial moles of HNO2 - Moles NaOH = Moles HNO2

Initial moles HNO2:

0.0200L * (0.10mol/L) = 0.00200moles HNO2

Moles NaOH = Moles NaNO2:

0.0100L * (0.10mol/L) = 0.00100 moles NaOH = Moles NaNO2

Moles HNO2:

0.00200moles - 0.00100moles = 0.00100 moles HNO2

Then, the added HCl reacts with NaNO2 producing HNO2:

HCl + NaNO2 → HNO2 + NaCl

Where moles added of HCl are moles of HNO2 produced and initial moles  of NaNO2 - moles of HCl = Moles NaNO2

Moles HCl:

4x10⁻⁵L * (10.0mol / L) = 0.00040moles HCl = Moles of HNO2 produced

Moles HNO2:

0.00100moles + 0.0004moles = 0.0014moles HNO2

Moles NaNO2:

0.00100moles - 0.0004moles = 0.0006moles NaNO2

With these moles and H-H equation we can solve for the pH of the buffer:

pH = pKa + log [NaNO2] / [HNO2]

Where pH is pH of the buffer

pKa is pKa of HNO2: 3.35

[] are moles of each species:

Replacing:

pH = 3.35 + log [0.0006moles NaNO2] / [0.0014moles HNO2]

pH = 2.98

Because a model would make the investigation easier to understand

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.