Question 4 of 32 Map o University Science Books presented by Sapling Learning er

Question

Question 4 of 32 Map o University Science Books presented by Sapling Learning eral Chemist Donald McQuarrie. Peter A. Rock Ethan Galloghy Calculate the change in pH when 9.00 mL of 0.100 M HCI(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4CI(aq). Number Calculate the change in pH when 9.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution. dlick to edit Number -O Previous ) Give Up & View Solution e Check Answer 0 Next Exit Y Hint For a buffer, pH-pK,+log [base] acid Note that this formula also works with moles (or millimoles) used in place of the concentrations.

Explanation / Answer

Solution :-

Lets first calculate the initial pH of the solution

pH = pka + log [base/acid]

pH= 9.26 + log [0.100/0.100]

pH = 9.26

now lets calculate the pH when 9 ml 0.100 M HCl is added

lets first calculate the moles of the each species

moles = molarity * volume in liter

moles of NH3 = 0.100 mol per L * 0.100 L = 0.01 mol

moles of NH4^+ = 0.100 mol per L * 0.100 L = 0.01 mol

moles of HCl = 0.100 mol per L * 0.009 L =0.0009 mol HCl

after reaction NH3 will convert to NH4^+

so new moles of NH3 = 0.01 mol – 0.0009 mol =0.0091 mol

moles of NH4^+ = 0.01 +0.0009 mol = 0.0109 mol

now lets calculate their molarity at total volume (100 +9 = 109 ml = 0.109 L)

new molarity NH3 = 0.0091 mol / 0.109 L = 0.083486 M

new molarity of the NH4+ = 0.0109 mol / 0.109 L = 0.1 M

now ;lets calculate ethe pH

pH= pka + log [base/ acid]

pH= 9.26+ log[0.083486/0.1]

pH= 9.18

therefore delta pH = 9.18-9.26 = -0.080

now lets calculate the delta pH when NaOH is added 0.100 M and 9.0 ml

moles of NH3 = 0.01 mol

moles of NH4+ = 0.01 mol

moles of NaOH = 0.100 mol per L * 0.009 L = 0.0009 mol NaOH

after reaction NaOH reacts with NH4+ so new moles are as follows

new moles of NH3 = 0.01 +0.0009 = 0.0109 mol

moles of NH4+ = 0.01 – 0.0009 = 0.0091 mol

new molarity at total volume are as follows

[NH3] = 0.0109 mol /0.109 L= 0.1 M

[NH4+] = 0.0091 mol / 0.109 L = 0.0834862 M

Now lets calculate the pH

pH= pka + log [base/acid]

pH= 9.26 + log [0.1/0.083486]

pH= 9.34

delta pH = 9.34 – 9.26 = 0.080

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