Question

A) How many moles of sodium hydroxide would have to be added to 225 mL of a 0.362 M hydrocyanic acid solution, in order to prepare a buffer with a pH of 9.580?

B) An aqueous solution contains 0.383 M nitrous acid.  

How many mL of 0.287 M sodium hydroxide would have to be added to 250 mL of this solution in order to prepare a buffer with a pH of 3.600?

C) An aqueous solution contains 0.313 M ethylamine (C₂H₅NH₂).  

How many mL of 0.223 M hydrochloric acid would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 10.400?

Answer 1

A)

Here [H⁺] = 0.362M

pH= -log [H⁺] = -log(0.362) = 0.4412

The pH of the buffer= 9.580

So pOH= 14-9.580= 4.42

So Final [OH]= 3.80*10^-5M

Let say xM NaOH required to make the pH at 9.580

Now x-0.362M = 3.80*10^-5

or x = 0.362+ 3.80*10^-5 = 0.362038M

Hence 0.362038M amount of NaOH is required to make 9.580 pH buffer.

[Note: I have done it by taking 1000ml solution. As the question is how many moles of NaOH , we cant say it precisely as the volume is not fixed. If the Volume of NaOH would be fixed then only the precise value of No of mole can be calculated.

B) 250ml of 0.383M HNO3 = 95.75mM

x ml of 0.287M of NaOH= 0.287xmM

Total Volume of Solution = 250+x

resulting solution has pH= 3.6

Hence [H⁺] = 10^-3.6= 2.5*10^-4

So (95.75 - 0.287x) / (250+x) = 2.5*10^-4

^{x= 333.11ml}

^{Hence 333.11 ml of NaOH should be added.}