How many times is the $\left[ {{H}^{+}} \right]$ in the blood ($pH$=7.36) greater than in spinal fluid ($pH$=7.53)?
(A) 1 times
(B) 1.5 times
(C) 2.0 times
(D) 2.5 times
Answer
644.1k+ views
Hint: The comparison between two values of $pH$ is asked. Therefore, we have to take the ratio between them. Keep in mind that $pH$ is nothing but the negative logarithm of hydrogen ion concentration and $pOH$ is the negative logarithm of hydroxide ion concentration.
Complete step by step solution:
-As we know the term acid and base have been defined in different ways in chemistry. Generally, we can say that an acid is any hydrogen-containing substance which is capable of donating hydrogen ion(proton) to another substance.
-Solutions can be classified as acidic or basic on their hydrogen ion concentration relative to pure water. If the $pH$ is below seven, then the solution is acidic and if it is greater than seven the solution is said to be basic.
- The concept of $pH$can be defined as the potential hydrogen ion concentration. That is, $pH$is the negative logarithm of $\left[ {{H}^{+}} \right]$ and in a similar way $pOH$is the negative logarithm of $\left[ O{{H}^{-}} \right]$
- Since the given $pH$belongs to alkaline region, we are going to use in terms of $pOH$
We know that,
$pH+pOH=$14
$pOH$=14−$pH$
Therefore, we could find the $pOH$of both fluids from $pH$ by this relation as follows
For blood, $pOH$=14−7.36=6.64
For spinal fluid,$pOH$= 14−7.53=6.47
As we mentioned above $pOH$is the negative logarithm of $\left[ O{{H}^{-}} \right]$
$pOH$=$-{{\log }_{10}}\left[ O{{H}^{-}} \right]$
$\left[ O{{H}^{-}} \right]$= Antilog $\left[ O{{H}^{-}} \right]$
Hence by using the above relation we could find $\left[ O{{H}^{-}} \right]$for blood and spinal fluid as follows
For blood, $\left[ O{{H}^{-}} \right]$=Antilog (−6.64) = $2.29\times {{10}^{-7}}mol/d{{m}^{3}}$
For spinal fluid, $\left[ O{{H}^{-}} \right]$=Antilog (−6.47) = $3.39\times {{10}^{-7}}mol/d{{m}^{3}}$
We can use the following equation which can be derived from the concept of
$pOH$ = 14 − $pH$,
for finding the hydrogen ion concentration. That is,
$\left[ {{H}^{+}} \right]$=$\left[ O{{H}^{-}} \right]$
Hence, for finding the ratio of $\left[ {{H}^{+}} \right]$we could write as follows
$\dfrac{\left[ {{H}^{+}} \right]for\text{ }blood}{\left[ {{H}^{+}} \right]~for\text{ }spinal\text{ }fluid}=\dfrac{{{10}^{-14}}}{2.29\times {{10}^{-7}}}\times \dfrac{3.39\times {{10}^{-7}}}{{{10}^{-14}}}$ = 1.5
Therefore, the answer is option(B) 1.5 times.
Note: The answer can also be found in an easy way.
$pH$for blood =7.36
We can find the hydrogen ion concentration from $pH$as follows
$\left[ {{H}^{+}} \right]$=${{10}^{-7.36}}$M
$pH$ for spinal fluid=7.53
$\left[ {{H}^{+}} \right]$=${{10}^{-7.53}}$M
Therefore,
$\dfrac{\left[ {{H}^{+}} \right]for\text{ }blood}{\left[ {{H}^{+}} \right]~for\text{ }spinal\text{ }fluid}=\dfrac{{{10}^{-7.36}}}{{{10}^{-7.53}}}={{10}^{7.56-7.36}}={{10}^{0.17}}$ =1.5
Complete step by step solution:
-As we know the term acid and base have been defined in different ways in chemistry. Generally, we can say that an acid is any hydrogen-containing substance which is capable of donating hydrogen ion(proton) to another substance.
-Solutions can be classified as acidic or basic on their hydrogen ion concentration relative to pure water. If the $pH$ is below seven, then the solution is acidic and if it is greater than seven the solution is said to be basic.
- The concept of $pH$can be defined as the potential hydrogen ion concentration. That is, $pH$is the negative logarithm of $\left[ {{H}^{+}} \right]$ and in a similar way $pOH$is the negative logarithm of $\left[ O{{H}^{-}} \right]$
- Since the given $pH$belongs to alkaline region, we are going to use in terms of $pOH$
We know that,
$pH+pOH=$14
$pOH$=14−$pH$
Therefore, we could find the $pOH$of both fluids from $pH$ by this relation as follows
For blood, $pOH$=14−7.36=6.64
For spinal fluid,$pOH$= 14−7.53=6.47
As we mentioned above $pOH$is the negative logarithm of $\left[ O{{H}^{-}} \right]$
$pOH$=$-{{\log }_{10}}\left[ O{{H}^{-}} \right]$
$\left[ O{{H}^{-}} \right]$= Antilog $\left[ O{{H}^{-}} \right]$
Hence by using the above relation we could find $\left[ O{{H}^{-}} \right]$for blood and spinal fluid as follows
For blood, $\left[ O{{H}^{-}} \right]$=Antilog (−6.64) = $2.29\times {{10}^{-7}}mol/d{{m}^{3}}$
For spinal fluid, $\left[ O{{H}^{-}} \right]$=Antilog (−6.47) = $3.39\times {{10}^{-7}}mol/d{{m}^{3}}$
We can use the following equation which can be derived from the concept of
$pOH$ = 14 − $pH$,
for finding the hydrogen ion concentration. That is,
$\left[ {{H}^{+}} \right]$=$\left[ O{{H}^{-}} \right]$
Hence, for finding the ratio of $\left[ {{H}^{+}} \right]$we could write as follows
$\dfrac{\left[ {{H}^{+}} \right]for\text{ }blood}{\left[ {{H}^{+}} \right]~for\text{ }spinal\text{ }fluid}=\dfrac{{{10}^{-14}}}{2.29\times {{10}^{-7}}}\times \dfrac{3.39\times {{10}^{-7}}}{{{10}^{-14}}}$ = 1.5
Therefore, the answer is option(B) 1.5 times.
Note: The answer can also be found in an easy way.
$pH$for blood =7.36
We can find the hydrogen ion concentration from $pH$as follows
$\left[ {{H}^{+}} \right]$=${{10}^{-7.36}}$M
$pH$ for spinal fluid=7.53
$\left[ {{H}^{+}} \right]$=${{10}^{-7.53}}$M
Therefore,
$\dfrac{\left[ {{H}^{+}} \right]for\text{ }blood}{\left[ {{H}^{+}} \right]~for\text{ }spinal\text{ }fluid}=\dfrac{{{10}^{-7.36}}}{{{10}^{-7.53}}}={{10}^{7.56-7.36}}={{10}^{0.17}}$ =1.5
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

