Question 11 Which of the following is the net ionic equation...

Question 11 Which of the following is the net ionic equation for: CaBr$_2$ (aq) + Na$_3$PO$_4$ (aq) ? ... 5 pts On your work sheet for this question write out the full reaction and identify the spectator ions. Use a solubility chart to determine solubility. 3 Br$_2$ (aq) + 6 Na$^+$ (aq) ? 6 NaBr (aq) 3 Ca$^{2+}$ (aq) + 2 PO$_4^{3-}$ (aq) ? Ca$_3$(PO$_4$)$_2$ (s) 6 Br$^-$ (aq) + 6 Na$^+$ (aq) ? 6 NaBr (aq) 3 Ca$^{2+}$ (aq) + 6 Br$^-$ (aq) + 6 Na$^+$ (aq) + 2 PO$_4^{3-}$ (aq) ? 6 Na$^+$ (aq) + 6 Br$^-$ (aq) + Ca$_3$(PO$_4$)$_2$ (s) 3 CaBr$_2$ (aq) + 2 Na$_3$PO$_4$ (aq) ? 6 NaBr (aq) + Ca$_3$(PO$_4$)$_2$ (s)

Transcript

00:02 Hi there.

00:03 In this question, we have a double replacement reaction, or your textbook may call them double displacement.

00:11 The two terms are used interchangeably.

00:13 It just depends on which textbook you're using.

00:16 But a double replacement and a double displacement are the same thing.

00:20 In this type of reaction, the ions exchange places.

00:24 So we are starting with strontium bromide, sr, br2, and it is aquius.

00:33 We're combining that with sodium phosphate, na3, po4, which is also aquas.

00:45 All right, so these are going to exchange ions.

00:48 The positive ion from the first goes with the negative ion from the second, and we need to write the formulas correctly.

00:58 Stronthium has a two positive charge, whereas phosphate has a three negative charge.

01:04 So we need three strontiums and two phosphates so that the total sum of the charges adds to zero.

01:11 Our other product then is going to be the sodium combining with the bromide.

01:18 And it's a positive one negative one, so it's just nabr.

01:22 Okay, looking at the rules for solubility, we see that any ionic compounds composed with a group 1a metal are soluble.

01:32 So the sodium bromide has to be soluble because the sodium's a group 1a metal.

01:38 However, we see that most phosphates, except for group 1a metals and ammonium, phosphates are going to be insoluble.

01:46 So this strontium phosphate is our solid.

01:50 Okay, the next thing we need to do is balance this.

01:53 So we see that there's one strontium on the reactant side, but three on the product side, meaning i need a coefficient of three.

02:00 Three times two gives me six bromides, so coming over here to the product side, i need a six in front of nabr to give me six bromides.

02:09 Now i have six sodiums on the product side and three on the reactant side.

02:14 So i'm 82 in front of na3po4.

02:18 That gives me two phosphates and we have two phosphates on the product side.

02:22 Okay, so we've created the total balanced equation and now we need to write the net ionic equation because that's what the quite answer is asking for is the net ionic equation.

02:36 The net ionic equation shows the ions that combine to produce the precipitate or the solid...

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