Question

How many ml of 0.112 M Pb(NO3)2 are needed to completely react with 10.0ml of a 0.105 M KI Given Pb(NO3)2 (aq) + 2KI(aq)=2KNO3(aq)?

Answer 1

0.105 x 10 /2 / 0.112= 4.6875 ml of 0.112M Pb(NO3)2

Answer 2

Answer : The volume of
`Pb(NO_3)_2` needed are, 4.6875 ml

Explanation :

First we have to calculate the moles of
`KI`.

`\text{Moles of }KI=\text{Molarity of }KI* \text{Volume of solution}=0.105mole/L* 0.01L=0.00105mole`

Now we have to calculate the moles of
`Pb(NO_3)_2`

The given balanced chemical reaction is,

`Pb(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+PbI_2`

From the balanced chemical reaction, we conclude that

As, 2 moles of KI react with 1 mole of
`Pb(NO_3)_2`

So, 0.00105 moles of KI react with
`(0.00105)/(2)=0.000525` mole of
`Pb(NO_3)_2`

Now we have to calculate the volume of
`Pb(NO_3)_2`

`\text{Volume of }Pb(NO_3)_2=\frac{\text{Moles of }Pb(NO_3)_2}{\text{Molarity of }Pb(NO_3)_2}`

`\text{Volume of }Pb(NO_3)_2=(0.000525mole)/(0.112mole/L)=4.6875* 10^(-3)L=4.6875ml`

conversion used : (1 L = 1000 ml)

Therefore, the volume of
`Pb(NO_3)_2` needed are, 4.6875 ml