Question

# How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be...

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 550. mL of 0.976-M solution of NH3 in order to prepare a pH = 9.15 buffer?

What volume (to the nearest 0.1 mL) of 4.00-M NaOH must be added to 0.600 L of 0.300-M HNO2 to prepare a pH = 3.50 buffer?

first of all pKb of NH3 should be given which is needed.

i am writing all the steps ann so please use the Pkb and calculate the answer

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pOH = pKb + log(salt/base)

required pH = 9.15 so pOH = 14 -pH = 14-9.15 = 4.85

millimoles of NH3 = 550x.976 = 536.8 gram

let the NH4Cl mass added = x   => milliMoles of NH4Cl = (x/53.5 )x1000 =18.69x

so subsitiue all values

pOH = pKb + log(salt/base)

4.85 = pKb + log (18.69x/536.8)

solve it for x mass of the salt

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its acidic buffer so we need the pKa of the weak acid HNO2 its not given in the question

HNO2            +             NaOH -------> NaNO2   + H2O

.6x.3 moles                  4xv

180 millimoles                4v millimoles

after reaction Millmoles of salt = 4v

millimoles of Acid =    180-4v

pH = pKa + log (salt/acid)

3.5 = pka + log{ (4v/(180-4v)}

substitute the value of pka from the book, and find v the ml of NaOH needed

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