Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\begin{array}{l}{\text { Find the pH of } 0.004 \mathrm{M}\mathrm{\: NH}_{4} \mathrm{OH} \text { having } 3.2 \%} \\ {\text { dissociation. }}\end{array}" src="http://entrancecorner.oncod

\begin{array}{l}{\text { Find the pH of } 0.004 \mathrm{M}\mathrm{\: NH}_{4} \mathrm{OH} \text { having } 3.2 \%} \\ {\text { dissociation. }}\end{array}

Option: 1

-3.8894


Option: 2

38.894


Option: 3

4.55


Option: 4

3.8894


Answers (1)

best_answer

Given,

\begin{array}{l}{\mathrm{NH}_{4} \mathrm{OH} \rightarrow \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \text {(weak base) }} \\ {\mathrm{\: c} \quad\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \quad \: \: \: \: \: \: \: \: \: \: 0 \text {\: \: \: \: \: \: \: \: \: \: \: \: \: \: \ conc. before }} {\text { ionization }} \\ {\text { c(1-a) \: \: \: \: c.a \: \: \: \: \:c.a \: \: \: \: \:\: \: \: \: \: \: \: \: conc. after\: ionization }} \\ {\left[\mathrm{H}^{+}\right]=\mathrm{c} \cdot a=4 \times 10^{-3} \times \frac{3.2}{100}} \\\\ {\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left[1.28 \times 10^{-2}\right]} \\ {\mathrm{pH}= 1.8927 + 2\approx 3.8894}\end{array}
Therefore,option(4) is correct

Posted by

Gunjita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question