Question

Calculate the pH of (a) 4.2 x 10 M H SO, b)
0.006M NaOH c) 9.2 ~ 10 4M HNO3.​

Answer 1

To find:

pH of

• 4.2 × 10^(-4) H2SO4
• 0.006 M NaOH
• 9.2 × 10^(-4) HNO3

Calculation:

1) 4.2 × 10^(-4) M H2SO4

$pH = - log \{{H}^{ + } \}$

$= > pH = - log \{2 \times (4.2 \times {10}^{ - 4} ) \}$

$= > pH = - log \{8.4 \times {10}^{ - 4} \}$

$= > pH = 4 - log \{8.4 \}$

$= > pH = 4 - 0.93$

$\boxed{ = > pH = 3.07}$

2) 0.006 M NaOH

$p(OH) = - log \{ {OH}^{ - 1} \}$

$= > p(OH) = - log \{0.006 \}$

$= > p(OH) = - log \{6 \times {10}^{ - 3} \}$

$= > p(OH) = 3- log \{6 \}$

$= > p(OH) = 2.22$

$= > p(H) = 14 - 2.22$

$\boxed{ = > p(H) = 11.77}$

3) 9.2 × 10^(-4) M HNO3:

$pH = - log \{{H}^{ + } \}$

$= > pH = - log \{9.2 \times {10}^{ - 4} \}$

$= > pH = 4 - log \{9.2 \}$

$= > pH = 4 - 0.96$

$\boxed{ = > pH = 3.04}$

Hope It Helps.