Which ions remain in solution, unreacted, after each of the following pairs of solutions is mixed? (a) potassium carbonate and magnesium sulfate

Video transcript

everyone to hear as to identify the net ionic equation for the reaction of lead to nitrate and sodium iodide of PB you know three to Aquarius N. A. I. And this is gonna be a double displacement reaction because we have two compounds reacting. So they can switch ions to form new products. We're gonna look at each reactant if P. B. You know 3 2 chris is going to associate into PB. Two plus that's two And all -3 N. Ai and associate into N. A. Plus and I'm minus. So now we're gonna switch the ions and have PB. Two plus rafting with I minus. They give us PB A tube and it's a solid because salt of iodine are soluble except when paired with silver led to and mercury too. And then if you switch again we're gonna have an A. Plus An n. 0. 3 -. It's gonna form in a. n. 0. 3 chris. And it's gonna be soluble because salt of group one elements are soluble and stadium is in group one. We're gonna get PB. You know three to chris last two in A. I digress. It's gonna be a P. B. R. Two solid plus two and eight and +03 a twist. And now before writing the net ionic equation we need to write the total ionic equation first by writing all the ions and leaving any solid liquid or gas together. Do not break them up. We're gonna have PB two plus eight Plus 2 3 last year and a plus grades Plus two I - Chris. It's gonna get PB. I. too solid, That's two A plus a quiz last year In our 3 -8 grades. And for the net Ionic equation we can get rid of any spectator ions which exist as a reactant and product are spectator ions, R. N. A. Plus In n. 0. 3 -. So they're gonna cancel out for the net Ionic equation. We get PB two plus Plus two R -8 years. Just give us P. B. I. Two solid. Thanks for watching my video and I hope it was helpful.