Question

How many moles of K+ and PO4^3- ions are present in 20.0 mL of a 0.015 M solution of potassium phosphate?

Answer 1

`n_(K^+)=0.0009molK^+`

Step-by-step explanation:

Hello,

In this case, the first step is to compute the number of moles of potassium phosphate in 20.0 mL (0.020L) of the 0.015-M (mol/L) solution as shown below:

`n=0.020L*0.015(mol)/(L)=0.0003mol`

Thus, these moles correspond to potassium phosphate moles, which molecular formula is K₃PO₄, therefore, one mole of this compound contains three moles of potassium ions as it has three as its subscript in the formula. Thereby, the moles of potassium ions result in:

`n_(K^+)=0.0003molK_3PO_4*(3molK^+)/(1molK_3PO_4) \\\\n_(K^+)=0.0009molK^+`

Best regards.