600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M . The pH of the mixture is _____ . (Nearest Integer) [Given log 2 = 0.30 log 3 = 0.48 log 5 = 0.69 log 7 = 0.84 log 11 = 1.04]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M $H_{2} {SO}_{4}$ . The pH of the mixture is _____ $\times 10^{- 2}$ . (Nearest Integer)

[Given log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04] [2023]

Ans.
(186)

$\begin{matrix} HCl & ⟶ & H^{+} & + & {Cl}^{-} \\ Milli moles & 0.01 \times 600 & 6 \\ H_{2} {SO}_{4} & ⟶ & 2 H^{+} & + & {SO}_{4}^{- 2} \\ Milli moles & 0.01 \times 400 & 2 \times 4 = 8 \end{matrix}$

$[H^{+}] = \frac{8 + 6}{1000} = 14 \times 10^{- 3}$

   $pH = - \log (14 \times 10^{- 3})$

$= 3 - log 14$

    $= 3 - log (2×7)$

   $= 3-log 2-log 7$

   $= 3 - 0.30 - 0.84$

   $= 1.86 = 186 \times 10^{- 2}$

Want Us to Email you About Special Offers & Updates?