Chemistry, 04.04.2020 13:17 mahoganyking16
Calculate the [H+] in a solution that is 0.803 M in NaX and 0.677 M in HX given that the Ka of HX is 8.64 ⋅ 10 − 7 8.64⋅10-7. Report your answer in scientific notation to 3 sig figs.
Answers: 3
.![pH=pK_a+\log(\frac{[salt]}{[acid]})](/tpl/images/0582/4215/e4eea.png)
![pH=pK_a+\log(\frac{[NaX]}{[HX]})](/tpl/images/0582/4215/da324.png)
= Acid dissociation constant of HX = 
![pK_a=-\log[K_a]=-\log[8.64\times 10^{-7}]=6.06](/tpl/images/0582/4215/ec5c6.png)
![[salt]=[NaX]=0.803 M](/tpl/images/0582/4215/da1f0.png)
![[acid]=[HX]=0.677 M](/tpl/images/0582/4215/836f0.png)
![pH=-\log[H^+]](/tpl/images/0582/4215/cf945.png)
![6.14=-\log[H^+]](/tpl/images/0582/4215/0a3f8.png)
![[H^+]=10^{-6.14}=7.24\times 10^{-7}](/tpl/images/0582/4215/e51b5.png)
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Calculate the [H+] in a solution that is 0.803 M in NaX and 0.677 M in HX given that the Ka of HX is...
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