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
Part A 0.30 g of hydrogen chloride (HCl) is dissolved in water to make 2.5 L...
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