When #NH_4Cl# mixed with #NH_3# what happens?

Answer 1

You have mixed a weak acid with its conjugate base, and have thus prepared a BUFFER solution.

#NH_4^+(aq) + H_2O(l) rightleftharpoons NH_3(aq) + H_3O^+#
Now #K_a# #=# #([NH_3][H_3O^+])/([NH_4^+])#.
We can take #log_10# OF BOTH SIDES to give:
#log_10K_a# #=# #log_10[H_3O^+]# #+# #log_10(([NH_3])/([NH_4^+]))#.

Regarding reorganization:

#-log_10[H_3O^+]# #=# #-log_10K_a# #+# #log_10(([NH_3])/([NH_4^+]))#
But by definition, #-log_10[H_3O^+]# #=# #pH#, and, #-log_10K_a# #=# #pK_a#.
So, #pH# #=# #pK_a# #+# #log_10(([NH_3])/([NH_4^+]))#
So given concentrations of ammonia and ammonium chloride, the #pH# of the SOLUTION should remain TOLERABLY CLOSE to the #pK_a#, which from memory is #4.76#? When #[NH_3]# #=# #[NH_4^+]# then #pH# #=# #pK_a#, because #log_(10)1# #=# #0#.
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Answer 2

When NH₄Cl is mixed with NH₃, the NH₄Cl dissociates into NH₄⁺ and Cl⁻ ions. NH₃ acts as a base and accepts a proton from water to form NH₄⁺ and OH⁻ ions. The OH⁻ ions then react with the Cl⁻ ions to form water and Cl⁻ ions. So, the net result is the formation of NH₄⁺ ions, OH⁻ ions, and Cl⁻ ions in solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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