A25.0 ml solution of 0.100 m ch3cooh is titrated with a 0.200 m koh solution. calculate the ph after the following additions of the koh solution:

(a) 0.0 ml,
(b) 5.0 ml,
(c) 10.0 ml,
(d) 12.5 ml,
(e) 15.0 ml

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH =  0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-](1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH](3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

your answer would be a. 12.5 kg

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: