Answer:
[tex]M_{K^+}=2.1M[/tex]
Explanation:
Hello!
In this case, since we know the initial and final concentrations and the added volume of water, we can write:
[tex]3.50M*V_1=0.700M(0.210L+V_1)[/tex]
Which can be solved for the initial volume as follows:
[tex]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=\frac{0.147}{2.8}=0.0525L[/tex]
It means that the final volume is:
[tex]V_2=0.210L+0.0525L=0.2625L[/tex]
Next, we compute the moles of potassium phosphate in solution:
[tex]n_{K_3PO_4}=0.2625L*0.700\frac{molK_3PO_4}{L}=0.184molK_3PO_4[/tex]
Then, since 1 mol of potassium phosphate has 3 moles of potassium (K's subscript), we compute the moles of potassium ions:
[tex]n_{K^+}=0.184molK_3PO_4*\frac{3molK^+}{1molK_3PO_4}=0.551molK^+[/tex]
Finally, the concentration of potassium ions turns out:
[tex]M_{K^+}=\frac{0.551molK^+ }{0.2625L}\\\\ M_{K^+}=2.1M[/tex]
Regards!
