Question

(b) An unknown volume of 3.50 M potassium phosphate, K,PO, solution is added to 0.210 L of water

to form a 0.700 M K3PO, solution. Calculate the molarity of potassium ions, K in the solution. ​

Answer 1

`M_(K^+)=2.1M`

Step-by-step explanation:

Hello!

In this case, since we know the initial and final concentrations and the added volume of water, we can write:

`3.50M*V_1=0.700M(0.210L+V_1)`

Which can be solved for the initial volume as follows:

`3.50M*V_1-0.700M(0.210L+V_1)=0\\\\3.50V_1-0.147-0.7V_1=0\\\\2.8V_1=0.147\\\\V_1=(0.147)/(2.8)=0.0525L`

It means that the final volume is:

`V_2=0.210L+0.0525L=0.2625L`

Next, we compute the moles of potassium phosphate in solution:

`n_(K_3PO_4)=0.2625L*0.700(molK_3PO_4)/(L)=0.184molK_3PO_4`

Then, since 1 mol of potassium phosphate has 3 moles of potassium (K's subscript), we compute the moles of potassium ions:

`n_(K^+)=0.184molK_3PO_4*(3molK^+)/(1molK_3PO_4)=0.551molK^+`

Finally, the concentration of potassium ions turns out:

`M_(K^+)=(0.551molK^+ )/(0.2625L)\\\\ M_(K^+)=2.1M`

Regards!