
The $[O{H^ - }]$ in $100mL$ of $0.016M$ $HCl(aq)$ is:
(A) $6.25 \times {10^{ - 12}}M$
(B) $3 \times {10^{ - 10}}M$
(C) $6.25 \times {10^{ - 13}}M$
(D) $1.6 \times {10^{ - 3}}M$
Answer
466.5k+ views
Hint:In order to answer this question you must be aware of the concepts of the ionic equilibrium. And also recall the concept of finding pH of solutions. Firstly, Enlist all the given quantities and then find the no. of moles oh $[{H^ + }]$ ions present in the solution. And then use the formulae of pH of solutions to find the concentration of $[O{H^ - }]$ ions. Take care of the units and calculate with accuracy and then choose the correct answer.
Complete step-by-step solution:Step 1: In this step we will enlist all the given quantities:
Given concentration of $[HCl]$ = $0.016M$
Given volume of solution = $100mL$
Step 2: In this step, we will calculate the number of moles of $[{H^ + }]$ ions in the given solution:
No. of moles of $[{H^ + }]$ ions in $100mL$ solution = $\frac{{100 \times 0.016}}{{1000}}\,\, = \,\,$ $1.6 \times {10^{ - 3}}$
Step 3: In this step, we will use the concepts of pH to find the concentration of $[O{H^ - }]$ ions:
As, we know that,$[{H^ + }]$ $[O{H^ - }]$ = ${10^{ - 14}}$
$ \Rightarrow \,\,1000 \times \frac{{1.6 \times {{10}^{ - 3}}}}{{100}} \times [O{H^ - }]\, = \,\,{10^{ - 14}}$
$ \therefore \,\,[O{H^ - }]\,\, = \,\,$$6.25 \times {10^{ - 13}}M$
Hence we got the required concentration of $[O{H^ - }]$ ions = $6.25 \times {10^{ - 13}}M$
Clearly, the correct answer is Option (C)..
Note:Hydrogen ion concentration (the pH) is one of the important factors that affect growth and multiplication of algae and hence the oil and biodiesel production. Most algal growth occurs in the region of neutral pH, although optimum pH is the pH of initial culture in which an alga is adapted to grow. Found that pH of around 8 seems most beneficial for maximum growth rate and lipid accumulation of Nannochloropsis salina and to minimize invading organisms. However, adding buffers will not be cost-effective or realistic at a large scale.
Complete step-by-step solution:Step 1: In this step we will enlist all the given quantities:
Given concentration of $[HCl]$ = $0.016M$
Given volume of solution = $100mL$
Step 2: In this step, we will calculate the number of moles of $[{H^ + }]$ ions in the given solution:
No. of moles of $[{H^ + }]$ ions in $100mL$ solution = $\frac{{100 \times 0.016}}{{1000}}\,\, = \,\,$ $1.6 \times {10^{ - 3}}$
Step 3: In this step, we will use the concepts of pH to find the concentration of $[O{H^ - }]$ ions:
As, we know that,$[{H^ + }]$ $[O{H^ - }]$ = ${10^{ - 14}}$
$ \Rightarrow \,\,1000 \times \frac{{1.6 \times {{10}^{ - 3}}}}{{100}} \times [O{H^ - }]\, = \,\,{10^{ - 14}}$
$ \therefore \,\,[O{H^ - }]\,\, = \,\,$$6.25 \times {10^{ - 13}}M$
Hence we got the required concentration of $[O{H^ - }]$ ions = $6.25 \times {10^{ - 13}}M$
Clearly, the correct answer is Option (C)..
Note:Hydrogen ion concentration (the pH) is one of the important factors that affect growth and multiplication of algae and hence the oil and biodiesel production. Most algal growth occurs in the region of neutral pH, although optimum pH is the pH of initial culture in which an alga is adapted to grow. Found that pH of around 8 seems most beneficial for maximum growth rate and lipid accumulation of Nannochloropsis salina and to minimize invading organisms. However, adding buffers will not be cost-effective or realistic at a large scale.
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