Question

Part A 0.30 g of hydrogen chloride (HCl) is dissolved in water to make 2.5 L of solution. What is the pH of the resulting hydPart B 0.35 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 6.5 L of solution. What is the pH of this solPart A 0.10 M NH4Cl Express your answer to two decimal places. MO? PO AJO pH = 2.3 Submit Request Answer - Part B 0.17 M NaC,

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Answer #1

A) moles of HCl = mass/ molar mass = 0.30g/36.5g/mol

molarity = moles/ V(L) = [0.30/36.5] /2.5 = 0.003287 M

HCl being a strong acid

[acid] = [H+]

and pH = -log[H+] = -log 0.003287 = 2.4831

B) moles of NaOH = mass/molar mass = 0.35g/40g/mol

Volume = 6.5L

thus molarity of NaOH = [0.35/40] /6.5 = 0.001346 mol/L

As NaOH is a strong base [NaOH] = [OH-] and thus

pOH = -log [OH-] = -log 0.001346 = 2.87

pH = 14-pOh = 14-2.87 =11.13

2A)

0.10M NH4Cl

NH4Cl is a salt of weak base and strong acid , thus will be acidic with pH < 7

Its pH is calculated as

pH = 1/2[pKw-pKb -log C] and pKb of NH3 = 4.75

Thus pH = 1/2[ 14 -4.75 -log 0.1]=5.125

2B) 0.17M sodium acetate

C2H3O2H is a weak acid , so the salt of NaOH and aceetic acid makes a basic salt with pH > 7

pH of such salt = 1/2[pKw +pKa+ log C]

= 1/2 [14+4.75 + log 0.17]

= 8.99

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