Question

wame S tation Date CHM 112 - Acid-Base Titration using pH Sensor Pre-lab Questions 1. Calculate [H,O') and JOH) in solutions of the following pls. [H30") (OH) PH 4.18 5.70 8.56 2. A student titrated 20.00 mL of 0.410 M HCl with 0.320 M NaOH and collected the following data. Volume NaOH added, mL 0.00 0.39 0.46 4.00 0.54 6.00 0.62 8.00 0.70 10.00 0.78 12.00 0.87 14.00 0.96 16.00 1.07 18.00 20.00 1.35 22.00 1.56 24.00 1.93 24.50 2.09 25.00 235 25.50 3.06 26.00 11.40 26.50 11.80 27.00 12.00 28.00 12.20 29,00 12.30 1.19 a) Prepare a titration curve for this titration using Excel. Page 1 of 2

Answer 1

pH = -log[H⁺] = -log[H₃O⁺]

[H⁺] = [H₃O⁺] = 10^-pH

[OH-] = K_w / [H₃O⁺]

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1.

pH = 4.18 ==> [H₃O⁺] = 10^-4.18 = 0.0000661 = 6.61 x 10^-5 M

[OH-] = K_w / [H₃O⁺] = 1.0 x 10^-14 / 6.61 x 10^-5 M = 1.51 x 10^-10 M

[H₃O⁺] = 6.61 x 10^-5 M; [OH-] = 1.51 x 10^-10 M

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2.

pH = 5.70 ==> [H₃O⁺] = 10^-5.70 = 0.0000661 = 1.99 x 10^-5 M

[OH-] = K_w / [H₃O⁺] = 1.0 x 10^-14 / 1.99 x 10^-5 M = 5.01 x 10^-9 M

[H₃O⁺] = 1.99 x 10^-5 M; [OH-] = 5.01 x 10^-9 M

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3.

pH = 8.56 ==> [H₃O⁺] = 10^-8.56 = 0.0000661 = 2.75 x 10^-9 M

[OH-] = K_w / [H₃O⁺] = 1.0 x 10^-14 / 2.75 x 10^-9 M = 3.63 x 10^-6 M

[H₃O⁺] = 2.75 x 10^-9 M; [OH-] = 3.63 x 10^-6 M

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2a)

Chart Tools Design Home Insert Layout Format Page Layout R Bookl - Microsoft Excel Formulas Data Review View Ruler Formula Bar Q Gridlines Headings Zoom Message Bar Show/Hide 1 E = = split = Hide View Side by Side af Synchronous Scrolling 5 Macros Normal Page Page Break Custom Full Layout Preview Views Screen Workbook Views 100% Zoom to Selection Zoom New Window Arrange Freeze All Panes - Unhide Save Switch Workspace Windows Reset Window Position Window Macros с D | Ε G H K L M N O P Chart 1 в 1 Volume NaOH (mL)I PH 0.39 0.46 0.54 0.62 0.7 0.78 0.87 0.96 1.07 1.19 1.35 pH 1.56 pH 24 24.5 25 25.5 26 26.5 1.93 2.09 2.35 3.06 11.4 11.8 12 12.2 12.3 27 28 OR 29 0 5 10 25 30 35 15 20 Volume NaOH (mL) 25 26 Sheet1 Sheet2 Sheet3 Ready OU 100% +

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2b)

M1V1 = M2V2 is our formula

M1 = 0.410 M, V1 = 20.0 mL; M2 = 0.320 M; V2 = ?

V2 = M1V1 / M2 = (0.410 M x 20.0 mL) / 0.320 M = 25.625 mL

Volume of titrant required to reach the equivalence point = 25.625 mL

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2c)

moles [H⁺] = 20 mL x 10^-3 L x 0.410 M = 0.0082 mol

moles [OH-] = 21.50 mL x 10^-3 L x 0.320 M = 0.00688 mol

excess [H⁺] = 0.0082 mol - 0.00688 mol = 0.00132 mol

total volume = 20 + 21.5 = 41.5 mL = 0.0415 L

molarity = 0.00132 mol / 0.0415 L = 0.032 M

[H₃O⁺] = 0.032 M; [OH^-] = 3.14 x 10^-13; pH = 1.49

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