Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added to 28.0 mL of 0.230 M NaOH(aq) .

pH = 12.35

Explanation:

The HCl reacts with NaOH, thus:

HCl + NaOH → NaCl + H₂O

1 mole of HCl reacts per mole of NaOH

To know the pH we must know the moles of NaOH and the moles of HCl that reacts

Moles NaOH and HCl

NaOH: 0.0280L * (0.230mol / L) = 6.44x10⁻³ moles

HCl: 0.0230L * (0.230mol / L) = 5.29x10⁻³

As moles NaOH > Moles HCl, the moles that remains of NaOH are:

Moles NaOH = 6.44x10⁻³ - 5.29x10⁻³ = 1.15x10⁻³ moles NaOH = Moles OH⁻

In 23.0mL + 28.0mL = 51.0mL = 0.0510L:

1.15x10⁻³ moles OH⁻ / 0.0510L =

0.0225M OH⁻

pOH = -log [OH⁻]

pOH = 1.647

And pH = 14 - pOH

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

well in order to answer this question you have to provide more information < 3