pH calculation

[bharat008]

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I was trying to understand some buffer concepts. I know how to solve those questions using Handerson-H eqn.. I came across this one and couldn't solve it..

You have 100 mL of 1 M ammonia solution (pKa=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.5?

Answer is 36mL of HCl..

Here is the LINK on how they solved it but doesn't make sense to me.

 

[txlotusboys37]

...

 

[spoog74]

I was trying to understand some buffer concepts. I know how to solve those questions using Handerson-H eqn.. I came across this one and couldn't solve it..

You have 100 mL of 1 M ammonia solution (pKa=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.5?

Answer is 36mL of HCl..

Here is the LINK on how they solved it but doesn't make sense to me.

this question is pretty long so i doubt they'd put something like this on the DAT

 

[bharat008]

this question is pretty long so i doubt they'd put something like this on the DAT

That's what i thought.. But just wanted to know how it works out..

It would be great if anyone can explain it.

 

[RalphMouth]

So we know that from the HH equation, 9.5 = 9.25 + log (NH3)/(NH4+)

Therefore log (NH3/NH4) = 0.25, use a calculator to find that log x =0.25, where x = 1.78, the ratio of NH3/NH4+.

NH3 ratio to start is 0.1, so basically since HCl completely disassociates (and in a 1 H+ to HCl manner) we know that any HCl added will form the conjugate acid NH4+. If we represent the amount of HCl added as x, we know that NH3 will be 0.1 - x at the buffer solution (starting ammonia will react with the HCl added). We know that x will also be the amount of NH4+ formed.

Therefore we can represent NH3/NH4+ as (0.1-x)/x, we calculated this ratio as 1.78 earlier. Simply solve for x to get the amount of HCl added (also the amount of NH4+ formed).

 

[bharat008]

So we know that from the HH equation, 9.5 = 9.25 + log (NH3)/(NH4+)

Therefore log (NH3/NH4) = 0.25, use a calculator to find that log x =0.25, where x = 1.78, the ratio of NH3/NH4+.

NH3 ratio to start is 0.1, so basically since HCl completely disassociates (and in a 1 H+ to HCl manner) we know that any HCl added will form the conjugate acid NH4+. If we represent the amount of HCl added as x, we know that NH3 will be 0.1 - x at the buffer solution (starting ammonia will react with the HCl added). We know that x will also be the amount of NH4+ formed.

Therefore we can represent NH3/NH4+ as (0.1-x)/x, we calculated this ratio as 1.78 earlier. Simply solve for x to get the amount of HCl added (also the amount of NH4+ formed).

Thanks! Got it. This was so easy!! 👍