Answer to Question #99999 in Chemistry for Sarah

Question

Consider a titration between 40.0 mL of .200 M HNO3 with .100 M NaOH. Calculate the pH at specific points given throughout the titration.
A. No NaOH added
B. 10.0 mL NaOH
c. 40.0 mL NaOH
D. 80.0 mL NaOH
E. 100.0 mL NaOH

Answer 1

$$NaOH(aq)+ HNO_3(aq) =NaNO_3(aq) + H_2O(l)$$

A. No NaOH is added

$c(H^+)=c(HNO3) = 0.200 M$

$pH = -log_{10}[H^+]=-log_{10}(0.200) = 0.699$

B. 10 mL of NaOH are added

$n(HNO3) = c\times V = 0.200\times0.040= 0.008 mol$

$n(NaOH) = c\times V = 0.100\times 0.010 = 0.001 mol$

According to equation mole ratio $n(NaOH):n(HNO3) = 1:1$ , then moles of HNO3 used for reaction are $n(HNO3) = n(NaOH) = 0.001 mol$ , moles left after titration are $n(HNO_3)_{left} = 0.008-0.001 = 0.007 mol$

$c(H^+)= \frac{n}{V} = \frac{0.007}{0.04+0.01} = 0.14 M$

$pH = -log_{10}(0.14) = 0.854$

C. 40 .0 mL NaOH are added

$n(HNO3) = 0.008 mol$

$n(NaOH) = c\times V = 0.1\times 0.04 = 0.004 mol$

According to equation mole ratio n(NaOH):n(NaOH) , then moles of HNO3 used for reaction are n(HNO3) = n(NaOH) = 0.004 mol, moles left after titration are

$n(HNO_3)_{left} = 0.008-0.004 = 0.004 mol$

$c(H^+) = \frac{n}{V} = \frac{0.004}{0.04+0.04} = 0.05 M$

$pH = -log_{10}(0.05) = 1.3$

D. 80 mL of NaOH are added

$n(HNO3) = 0.008 mol$

$n(NaOH) = c\times V = 0.1\times 0.08=0.008 mol$

According to equation mole ratio n(NaOH):n(NaOH)=1:1 , then moles of HNO3 used for reaction are n(HNO3) = n(NaOH) = 0.008 mol, moles left after titration are

$n(HNO_3)_{left} = 0.008-0.008 = 0$

pH is determined by the $H^+$ ions produced by the water dissosiation

$[H^+] = 1\times 10^{-7} M$

$pH = -log_{10}(1\times10^{-7}) = 7$

E. 100 mL of NaOH are added

$n(HNO3) = 0.008 mol$

$n(NaOH) = 0.1\times 0.1 = 0.01 mol$

Moles of NaOH left:

$n(NaOH)_{left} = 0.01-0.008 = 0.002 mol$

$n(OH^-) = n(NaOH) = 0.002 mol$

$c(OH^-) = \frac{n}{V} = \frac{0.002} {0.04+0.1} = 0.0143 M$

$pOH = -log_{10}[OH^-] = -log_{10}(0.0143) = 1.845$

As pH + pOH =14, then pH = 14-1.845 = 12.15