First you need to write out the original equation.
![NiCO_{3(s)}+ HI_{aq} ->NiI_{2(aq)}+CO_{2(g)}+H_2O_{(l)}](https://tex.z-dn.net/?f=NiCO_%7B3%28s%29%7D%2B%20HI_%7Baq%7D%20-%3ENiI_%7B2%28aq%29%7D%2BCO_%7B2%28g%29%7D%2BH_2O_%7B%28l%29%7D)
It needs to be balanced, though, so after balancing it should look like this:
![NiCO_{3(s)}+ 2HI_{aq} ->NiI_{2(aq)}+CO_{2(g)}+H_2O_{(l)}](https://tex.z-dn.net/?f=NiCO_%7B3%28s%29%7D%2B%202HI_%7Baq%7D%20-%3ENiI_%7B2%28aq%29%7D%2BCO_%7B2%28g%29%7D%2BH_2O_%7B%28l%29%7D)
The ionic equation would look like this. (Only aqueous solutions can be split up).
![NiCO_{3(s)}+ 2H^+_{aq}+2I^-_{aq} ->Ni^{2+}_{aq}+2I^-_{(aq)}+CO_{2(g)}+H_2O_{(l)}](https://tex.z-dn.net/?f=NiCO_%7B3%28s%29%7D%2B%202H%5E%2B_%7Baq%7D%2B2I%5E-_%7Baq%7D%20-%3ENi%5E%7B2%2B%7D_%7Baq%7D%2B2I%5E-_%7B%28aq%29%7D%2BCO_%7B2%28g%29%7D%2BH_2O_%7B%28l%29%7D)
The net ionic equation only contains the products that are not aqueous. Therefore, the final net ionic equation should be:
![NiCO_{3(s)}+ 2H^+_{aq} ->Ni^{2+}_{aq}+CO_{2(g)}+H_2O_{(l)}](https://tex.z-dn.net/?f=NiCO_%7B3%28s%29%7D%2B%202H%5E%2B_%7Baq%7D%20-%3ENi%5E%7B2%2B%7D_%7Baq%7D%2BCO_%7B2%28g%29%7D%2BH_2O_%7B%28l%29%7D)
So your answer is C.
Answer:
Correct option : B
Explanation:
First compare
with ![OF_2](https://tex.z-dn.net/?f=OF_2)
there are two lone pair in
but only one lone pair presen in the
so more lone pair means more repulsion and hence less bond bond angle beacuse of lone pair repulsion bonded atom come closer and hence bond angle decreased .
bond angle inversely proporional to repulsion so more repulsion means less bond angle
![bond \, angle \, : OF_2 \textless PF_3](https://tex.z-dn.net/?f=bond%20%5C%2C%20angle%20%5C%2C%20%3A%20%20%20%20OF_2%20%5Ctextless%20PF_3)
does no exist it exist only in the form of
and here assuming
because in case of
difficult to compare bond angle because here two different type of bond angle is present.
![Compare \, PF_3 \, and \, PF_4^+](https://tex.z-dn.net/?f=Compare%20%5C%2C%20PF_3%20%5C%2C%20and%20%5C%2C%20PF_4%5E%2B)
there is no lone pair in the
so here will no be any repulsion and hence bond will be more in
as compared to ![PF_3](https://tex.z-dn.net/?f=PF_3)
![bond \, angle \, : PF_3 \textless PF_4^+](https://tex.z-dn.net/?f=bond%20%5C%2C%20angle%20%5C%2C%20%3A%20%20%20%20PF_3%20%5Ctextless%20PF_4%5E%2B)
![bond \, angle \, : OF_2 \textless PF_3 \textless PF_4^+](https://tex.z-dn.net/?f=bond%20%5C%2C%20angle%20%5C%2C%20%3A%20%20%20%20OF_2%20%5Ctextless%20PF_3%20%5Ctextless%20PF_4%5E%2B)
![Our \ goal \ here\ is \ to \ find \ an \ acid-base \ conjugate \ pair \ that \ have \a \\\$pK_a-p H$. This is because we assume that the acid-base conjugate pair have an equal concentration to be considered an effective buffer.](https://tex.z-dn.net/?f=Our%20%5C%20goal%20%5C%20here%5C%20%20is%20%5C%20to%20%5C%20find%20%5C%20an%20%5C%20acid-base%20%5C%20conjugate%20%5C%20pair%20%5C%20that%20%5C%20have%20%5Ca%20%5C%5C%5C%24pK_a-p%20H%24.%20This%20is%20because%20we%20assume%20that%20the%20acid-base%20conjugate%20pair%20have%20an%20equal%20concentration%20to%20be%20considered%20an%20effective%20buffer.)
![$$p H-p K_a+\log \frac{\left[A^{-}\right]}{[H A]}$$](https://tex.z-dn.net/?f=%24%24p%20H-p%20K_a%2B%5Clog%20%5Cfrac%7B%5Cleft%5BA%5E%7B-%7D%5Cright%5D%7D%7B%5BH%20A%5D%7D%24%24)
![$$Where $\left[A^{-}\right]-[H A]$, so $\log \frac{\left[A^{-}\right]}{[H A]}-\log 1-0_{\text {so ... }}$$$](https://tex.z-dn.net/?f=%24%24Where%20%24%5Cleft%5BA%5E%7B-%7D%5Cright%5D-%5BH%20A%5D%24%2C%20so%20%24%5Clog%20%5Cfrac%7B%5Cleft%5BA%5E%7B-%7D%5Cright%5D%7D%7B%5BH%20A%5D%7D-%5Clog%201-0_%7B%5Ctext%20%7Bso%20...%20%7D%7D%24%24%24)
pH-pKa
![$$Solve for $K_a$.$$](https://tex.z-dn.net/?f=%24%24Solve%20for%20%24K_a%24.%24%24)
![$$\begin{aligned}p H-p K_a &--\log K_a \\K_a &-10^{-p H} \\&-10^{-3.5} \\K_a &-3.2 \times 10^{-4}\end{aligned}$$](https://tex.z-dn.net/?f=%24%24%5Cbegin%7Baligned%7Dp%20H-p%20K_a%20%26--%5Clog%20K_a%20%5C%5CK_a%20%26-10%5E%7B-p%20H%7D%20%5C%5C%26-10%5E%7B-3.5%7D%20%5C%5CK_a%20%26-3.2%20%5Ctimes%2010%5E%7B-4%7D%5Cend%7Baligned%7D%24%24)
![$$The nearest pairs are $\mathrm{HCOCOOH} / \mathrm{HCOCOO}^{-}$with a $\mathrm{K}_{\mathrm{a}}-3.5 \times 10^{-4}$,](https://tex.z-dn.net/?f=%24%24The%20nearest%20pairs%20are%20%24%5Cmathrm%7BHCOCOOH%7D%20%2F%20%5Cmathrm%7BHCOCOO%7D%5E%7B-%7D%24with%20a%20%24%5Cmathrm%7BK%7D_%7B%5Cmathrm%7Ba%7D%7D-3.5%20%5Ctimes%2010%5E%7B-4%7D%24%2C)
![$\mathrm{HOCH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} / \mathrm{HOCH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{COO}^{-}$with a $\mathrm{K}_{\mathrm{a}}-2.9 \times 10^{-4}$, and](https://tex.z-dn.net/?f=%24%5Cmathrm%7BHOCH%7D_2%20%5Cmathrm%7BCH%7D%28%5Cmathrm%7BOH%7D%29%20%5Cmathrm%7BCOOH%7D%20%2F%20%5Cmathrm%7BHOCH%7D_2%20%5Cmathrm%7BCH%7D%28%5Cmathrm%7BOH%7D%29%20%5Cmathrm%7BCOO%7D%5E%7B-%7D%24with%20a%20%24%5Cmathrm%7BK%7D_%7B%5Cmathrm%7Ba%7D%7D-2.9%20%5Ctimes%2010%5E%7B-4%7D%24%2C%20and)
![$\mathrm{CH}_3 \mathrm{COOC}_6 \mathrm{H}_4 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COOC}_6 \mathrm{H}_4 \mathrm{COO}^{-}$with a $\mathrm{K}_a-3.6 \times 10^{-4}$. We can also check for the $\mathrm{K}_b$. First, solve for $p K_b$ then the $K_b$ -](https://tex.z-dn.net/?f=%24%5Cmathrm%7BCH%7D_3%20%5Cmathrm%7BCOOC%7D_6%20%5Cmathrm%7BH%7D_4%20%5Cmathrm%7BCOOH%7D%20%2F%20%5Cmathrm%7BCH%7D_3%20%5Cmathrm%7BCOOC%7D_6%20%5Cmathrm%7BH%7D_4%20%5Cmathrm%7BCOO%7D%5E%7B-%7D%24with%20a%20%24%5Cmathrm%7BK%7D_a-3.6%20%5Ctimes%2010%5E%7B-4%7D%24.%20We%20can%20also%20check%20for%20the%20%24%5Cmathrm%7BK%7D_b%24.%20First%2C%20solve%20for%20%24p%20K_b%24%20then%20the%20%24K_b%24%20-)
![$$\begin{aligned}p K_w &-p K_b+p K_a \\p K_b &-p K_w-p K_a \\&-14-3.5 \\p K_b &-10.5 \\p K_b &--\log K_b \\K_b &-10^{-p K_b} \\&-10^{-10.5} \\K_b &-3.2 \times 10^{-11}\end{aligned}$$](https://tex.z-dn.net/?f=%24%24%5Cbegin%7Baligned%7Dp%20K_w%20%26-p%20K_b%2Bp%20K_a%20%5C%5Cp%20K_b%20%26-p%20K_w-p%20K_a%20%5C%5C%26-14-3.5%20%5C%5Cp%20K_b%20%26-10.5%20%5C%5Cp%20K_b%20%26--%5Clog%20K_b%20%5C%5CK_b%20%26-10%5E%7B-p%20K_b%7D%20%5C%5C%26-10%5E%7B-10.5%7D%20%5C%5CK_b%20%26-3.2%20%5Ctimes%2010%5E%7B-11%7D%5Cend%7Baligned%7D%24%24)
![There \ are\ no \ pairs \ near \ this \ $K_b$ value.](https://tex.z-dn.net/?f=There%20%5C%20are%5C%20%20no%20%5C%20pairs%20%5C%20near%20%5C%20%20this%20%5C%20%20%24K_b%24%20value.)
<h3>What is a conjugate acid and base pair?</h3>
An acid-base pair that differs by one proton is referred to as a conjugate pair. A conjugate acid-base pair is a pair of substances that can both absorb and donate hydrogen ions to one another. A proton is added to the compound to create the conjugate acid, and a proton is taken out to create the conjugate base. When a proton is supplied to a base, a conjugate acid is created, and vice versa when a proton is taken away from an acid, a conjugate base is created.
To learn more about conjugate acid and base pair, visit;
brainly.com/question/10468518
#SPJ4