A sample of drinking water was tested for Pb2 and was found to have a Pb2 concentration of 17.50 ppb. A 9.00 mL sample of the drinking water was spiked with 1.00 mL of a 2.29 ppb Pb2 standard. Analysis of the spiked sample gave a concentration of 15.93 ppb Pb2 . Find the percent recovery of the spike.

Respuesta :

Answer:

[tex]X=75\%[/tex]

Explanation:

From the question we are told that:

Concentration [tex]C_1=17.50ppb[/tex]

Volume [tex]v=9mL[/tex]

Spike Volume [tex]V_s=1.00mL[/tex]

spike Conc [tex]C_2=2.29 ppb[/tex]

Analysis Conc [tex]C_s=15.93 ppb[/tex]

Generally the equation for  percent recovery is mathematically given by

 [tex]X=\frac{C_s'-C_1'}{C_2'}[/tex]

Where

Concentration of spiked sample C_s'

 [tex]C_s'=C_s*(v+v_s)[/tex]

 [tex]C_s'=15.92*(9+1)[/tex]

 [tex]C_2=159.2[/tex]

Concentration of unspiked sample C_1'

 [tex]C_1'=17.50*9mL[/tex]

 [tex]C_1'=157.5[/tex]

Concentration of spike sample C_2'

 [tex]C_2=2.29*1[/tex]

 [tex]C_2=2.29[/tex]

Therefore

 [tex]X=\frac{159.2-157.5}{2.29}[/tex]

 [tex]X=75\%[/tex]