Chemistry, asked by atharv2303, 11 months ago

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

Answers

Answered by nirman95
7

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.

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