Calculate the ph of the resulting solution if 31.0 ml of 0.310 m hcl (aq) is added to

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Katelynn Sweeney

Chemistry

25 May, 18:40

Calculate the ph of the resulting solution if 31.0 ml of 0.310 m hcl (aq) is added to

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Harper Fry · Answer 1

Missing question: 36.0 mL of 0.310 M NaOH (aq).

Chemical reaction: HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l).

Ionc net reaction: H⁺ (aq) + OH⁻ (aq) → H₂O (l)

c (HCl) = 0,310 M = 0,31 mol/L.

V (HCl) = 31,0 mL : 1000 mL/L = 0,031 L.

n (HCl) = c (HCl) · V (HCl).

n (HCl) = n (H⁺) = 0,31 mol/L · 0,031 L.

V (HCl) = 0,00961 mol.

n (NaOH) = 0,036 L · 0,31 mol/L.

n (NaOH) = n (OH⁻) = 0,01116 mol.

n (OH⁻) = 0,01116 mol - 0,00961 mol = 0,00155 mol.

c (OH⁻) = 0,00155 mol : (0,031 L + 0,036 L).

c (OH⁻) = 0,023 M.

pOH = - log (0,023 M) = 1,63.

pH = 14 - 1,63 = 12,37.