A 25.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of acetic acid is 4.5E-4
A 25.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of acetic acid is 4.5E-4
note that the actual value of Ka of acetic acid is 1.8*10^-5, not 4.5*10^-4
mmol = MV = 25*0.15 = 3.75 mmol of acetic acid
Vbase = 3.75/0.150 = 25 mL
total V = 25+25 = 50 mL
in equivalance:
CH3COO- + H2O <-> CH3COOH + OH-
Kb = [CH3COOH][OH-]/[CH3COO-]
Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.555*10^-10
5.555*10^-10 = x*x/(M-x)
M = mmol of CH3COO- / total V = 3.75 / 50 = 0.075 M
5.555*10^-10 = x*x/(0.075 -x)
X = [OH-] = 6.45*10^-6
pOH= -log(6.45*10^-6) = 5.190
pH = 14-5.190 = 8.81
A 25.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH...
A 25.0 mL sample of 0.150 M formic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of formic acid is 1.8 ⋅ 10-4.
A 50.0 ml sample of 0.50 M acetic acid, ch3cooh is titrated with a 0.150 M NaOH solution. calculate the ph after 25.0 ml of the base have been added (ka=1.8x10^-5)
A 25.0 mL sample of 0.150 M chloroacetic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The of chloroacetic acid is 1.4 ×10-3.
A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH after the addition of 13.0 mL of NaOH? The Ka of benzoic acid is 6.3x10-5.
8) A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH at the eq. point? pka= 4.20
1) A 25.0 mL sample of 0.150 M hydrazoic acid is titrated with a 0.150 M NaOH solution. What is the pH after 26.0 mL of base is added? The of hydrazoic acid is 1.9 × 10-5. 2)A 25.0 mL sample of 0.723 M HClO4 is titrated with a KOH solution. The H3O+ concentration after the addition of of KOH is ________ M.
A 25.0 mL sample of an acetic acid solution (Ka = 1.76 × 10-5) is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid is __________ M.
A student titrated a 100.0 mL sample of 0.100 M acetic acid with 0.050 M NaOH. (For acetic acid, Ka = 1.8 * 10^-5 at this temperature.) (a) Calculate the initial pH. (b) Calculate the pH after 50.0 mL of NaOH has been added. (c) Determine the volume of added base required to reach the equivalence point. (d) Determine the pH at the equivalence point?
A 20.0 mL sample of a 0.0875 M solution of acetic acid, CH3COOH, is titrated with a 0.115 M solution of KOH. What is the pH after 20.0 mL of KOH solution have been added? For CH3COOH, Ka=1.8 x 10–5.
A 25.0 mL sample of 0.293 M acetic acid solution is titrated with 37.5 mL of 0.195 M NaOH solution. What is the pH of this solution if K = 1.8 x 10-5? Final Answer: _______