Question

What volume of 0.1025 M HCl must be added to 15.64 ml of 0.0956 M NH3 to produce a solution of ph= 9.00 ?

Answer #1

**we have formula pOH = pkb + log
[NH4Cl]/[NH3] whre pkb of NH3 is 4.75**

**given pH = 9 , hence pOH = 14-pH = 14-9 = 5**

**hence 5 = 4.75 + log [NH4Cl]/[NH3]**

**[NH4Cl] = 1.78 [NH3]**

**NH4Cl moles = 1.78 x NH3 moles ( Moles
= Molarity x vol , vol term from both sides cancells
out)**

**NH3 + HCl ---> NH4Cl is the
equation**

**Initial NH3 moles = M x V ( in L) = 0.0956 x 15.64/1000
= 0.001495**

**Let HCl moles added = m**

**then after reactiong with HCl moles of NH3 = 0.001495-m
, NH4Cl moles = m**

**we have relation NH4Cl moles = 1.78 x NH3
moles**

**hence m = 1.78 ( 0.001495-m)**

**m = 0.00266 - 1.78m**

**m = 0.0009573 = HCl moles**

**now we have formula Molarity of HCl = moles
/vol**

**0.1025 = 0.0009573 /vol**

**vol = 0.00933 L = 9.33 ml = 9.3 ml ( rounded
value)**

a) What volume (to the nearest 0.1 mL) of 6.60-M HCl must be
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a pH = 7.40 buffer?
b) How many grams (to the nearest 0.01 g) of NH4Cl
(Mm = 53.49 g/mol) must be added to 400. mL of 0.964-M
solution of NH3 in order to prepare a pH = 8.75 buffer?
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What mass of ammonium chloride must be added to exactly 500 ml
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