Notice that it automatically checks because the total of B' is correctly 0.7: B B' Totals A 0.1 0.4 0.5 A' 0.2 0.3 0.5 Totals 0.3 0.7 1.0 Now we can find any of the 8 possible conditional probabilities: In particular, we want to find P(A given B), since B is given, we ignore everything except the B column, the. Simple and best practice solution for a/-2.35-5.6=-6.8 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.
A 35.0mL sample of 0.150 M acetic is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added?a) 0 mL
b) 17.5 mL
c) 34.5 mL
d) 35.0 mL
e) 35.5 mL
f) 50.0 mL
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WARNING! Very long answer! The pH values are a) 2.79; b) 4.75; c) 6.59; d) 8.81;
e) 11.03; f)12.42.
a) 0 mL
#color(white)(mmmmmmml)'HA' + 'H'_2'O' ⇌ 'A'^'-' + 'H'_3'O'^'+'#
#'I/mol·L'^'-1':color(white)(mm)0.150color(white)(mmmmml)0color(white)(mmm)0#
#'C/mol·L'^'-1':color(white)(mmll)'-'xcolor(white)(mmmmml)'+'xcolor(white)(mm)'+'x#
#'E/mol·L'^'-1':color(white)(l)0.150 - xcolor(white)(mmmmll)xcolor(white)(mmm)x#
#K_'a' = (['H'_3'O'^+]['A'^'-'])/(['HA']) = x^2/(0.150-x) = 1.76 × 10^-5#
#0.150/(1.76 × 10^'-5') = 8523 > 400#. ∴ #x ≪ 0.150##
Ulysses 2 7. #x^2 = 0.150 × 1.76 × 10^'-5' = 2.64 × 10^'-6'#
#x = sqrt(2.64 × 10^'-6') = 1.62 × 10^'-3'#
#['H'_3'O'^'+'] = xcolor(white)(l) 'mol/L' = 1.62 × 10^'-3'color(white)(l) 'mol/L'#
#'pH' = -log['H'_3'O'^'+'] = -log(1.62 × 10^'-3') = 2.79#
b) At 17.5 mL
This is the point of half-neutralization.
;:#'pH' = pK_'a' = -log(1.76 × 10^'-5') = 4.75#
A Text 2 35 5 0 Cmc) At 34.5 mL
#color(white)(mmmmmm)'HA' +color(white)(m) 'OH'^'-'color(white)(m) ⇌ color(white)(m)'A'^'-' + 'H'_2'O'#
#'I/mmol':color(white)(ml)5.25color(white)(mml)5.175color(white)(mmmll)0#
#'C/mmol':color(white)(ll)'-5.175'color(white)(ml)'-5.175'color(white)(mm)'+5.175'#
#'E/mmol':color(white)(m)0.075color(white)(mm)0color(white)(mmmmll)5.175#
#'Initial moles of HA' = 0.0350 color(red)(cancel(color(black)('L'))) × '0.150 mol'/(1 color(red)(cancel(color(black)('L')))) = '0.005 25 mol' = color(red)(bb'5.25 mmol')#
#'Moles of NaOH added' = 0.0345 color(red)(cancel(color(black)('L'))) × '0.150 mol'/(1 color(red)(cancel(color(black)('L')))) = '0.005 175 mol' = '5.175 mmol'#
#'Moles of HA remaining' = ('5.25 -5.175) mmol' = '0.075 mmol'#
#'pH' = 'p'K_'a' + log((['A'^'-'])/(['HA'])) = 4.75 + log((5.175 color(red)(cancel(color(black)('mol'))))/(0.075 color(red)(cancel(color(black)('mol'))))) = 4.75 + log(69.0) = 4.75 + 1.84 = 6.59#
d) At 35.0 mL
#color(white)(mmmmmm)'HA' +color(white)(m) 'OH'^'-'color(white)(m) ⇌ color(white)(m)'A'^'-' + 'H'_2'O'#
#'I/mmol':color(white)(ml)5.25color(white)(mml)5.25color(white)(mmmml)0#
#'C/mmol':color(white)(ll)'-5.25'color(white)(mm)'-5.25'color(white)(mlmm)'+5.25'#
#'E/mmol':color(white)(m)0color(white)(mmmm)0color(white)(mmmmmll)5.25#
You have a solution of 5.25 mmol of #'A'^'-'# in 70.0 mL.
#['A'^'-'] = '5.25 mmol'/'70.0 mL' = '0.0750 mol/L'#
#color(white)(mmmmmmm)'A'^'-' +color(white)(m) 'H'_2'O'color(white)(m) ⇌ color(white)(m)'HA' + 'OH'^'-'#
#'I/mmmol':color(white)(mml)0.0750color(white)(mmmmmmmmll)0color(white)(mml)0#
#'C/mmol':color(white)(mmmm)'-'xcolor(white)(mmmmmmmmm)'+'xcolor(white)(mll)'+'x#
#'E/mmol':color(white)(mml)0.0750 - xcolor(white)(mmmmmmml)xcolor(white)(mml)x#
#K_'b' = K_'w'/K_'a' = (1.00 × 10^'-14')/(1.76 × 10^'-5') = 5.68 × 10^'-10'# https://herejfiles887.weebly.com/sony-ip-camera-viewer-software.html.
#K_'b' = (['HA']['OH'^'-'])/(['A'^'-']) = x^2/(0.0750-x) = 5.68 × 10^'-10'#
A Text 2 35 5 000#0.0750/(5.68×10^'-10') = 1.32 × 10^8 > 400#. ∴ #x ≪ 0.0750#
#x^2 = 0.0750 × 5.68 × 10^'-10' = 4.26 × 10^'-11'# Sidify music converter 1 1 7 download free.
#x = sqrt(4.26 × 10^'-11') = 6.53 × 10^'-6'# Beatunes 4 5 6 download free.
#['OH'^'-'] = 6.53 × 10^'-6' color(white)(l)'mol/L'#
#'pOH' = -log(6.53 × 10^'-6') = 5.19#
#'pH' = '14.00 - pOH' = 14.00 - 5.19 = 8.81#
e) At 35.5 mL
You are now adding excess moles of #'OH'^'-'#
#'Excess moles of OH'^'-' = 0.5 color(red)(cancel(color(black)('mL OH'^'-'))) × '0.150 mmol OH'^'-'/(1 color(red)(cancel(color(black)('mL OH'^'-')))) = '0.075 mmol OH'^'-'#
#['OH'^'-'] = '0.075 mmol'/'70.5 mL' = 1.06 × 10^'-3' 'mol/L'#
#'pOH' = -log(1.06 × 10^'-3' ) = 2.97#
#'pH' = '14.00 - 2.97' = 11.03#
A Text 2 35 5 00f) At 50.00 mL
Airmail 2 6 mas download free. #'Excess moles of OH'^'-' = 15.0 color(red)(cancel(color(black)('mL OH'^'-'))) × '0.150 mmol OH'^'-'/(1 color(red)(cancel(color(black)('mL OH'^'-')))) = '2.25 mmol OH'^'-'#
#['OH'^'-'] = '2.25 mmol'/'85.0 mL' = '0.0265 mol/L'#
#'pOH' = -log(0.0265) = 1.58#
A Text 2 35 5 0 M#'pH' = '14.00 - 1.58' = 12.42#
Your calculated values should match the titration curve below. Alien skin blow up 3 1 4 276.
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