Question

Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

Answer 1

according to Arrhenius theory,
$\quad\bf{pH=-log[H^+]}$

where $H^+$ is the concentration of hydrogen ion.

(a) 0.003M HCl ,
HCl dissociate in .. $HCl\rightarrow H^++Cl^-$

so, concentration of $H^+$ = 0.003M

now, pH = -log(0.003) = -log(3 × 10^-3)

pH = 3 - log3 = 3 - 0.477 = 2.53

(b) 0.005M NaOH,
dissociation of NaOH is... $NaOH\rightarrow Na^++OH^-$

concentration of $OH^-$ = 0.005M

now, pOH = -log(0.005) = -log(5 × 10^-3)

pOH = 3 - log5 = 3 - 0.6989 = 2.3011

so, pH = 14 - pOH = 14 - 2.3011 = 11.6989

(c) 0.002M HBr,
dissociation of HBr is ... $HBr\rightarrow H^++Br^-$

concentration of $H^+$ = 0.002M

now, pH = -log(0.002) = -log(2 × 10^-3)

pH = 3 - log2 = 2.6987

(d) 0.002M KOH,
KOH dissociate in .. $KOH\rightarrow K^++OH^-$

so, concentration of $OH^-$ = 0.002M

so, pOH = -log(0.002) = -log(2 × 10^-3)

pOH = 3 - log2 = 2.6987

so, pH = 14 - 2.6987 = 11.30103

Answer 2

Answer:

1

ph=-log(h3o+)

= -(log 3 -3log 10)

=-(o. 4771 -3)

=2.5322

Explanation:

1

log 3=4771&log10=1