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twalters88
31.08.2020 •
Chemistry
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) .
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Ответ:
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
pH = 12.35Ответ:
Answer : The metal is copper.
Explanation :
As we are given that:
Mass of metal = 25 g
Volume of metal = 40.0 mL
Formula used:
Now putting all the given values in this formula, we get:
From this we conclude that the metal is copper whose density is 8.86 g/mL.
Hence, the metal is copper.