NOTE ON pH AND pOH CALCULATION

  • The *pH or pOH of a solution* is defined as the degree of acidity or alkalinity of a given chemical solution.
        This means that a chemical solution maybe either acidic in character or alkaline(basic) or in between acidicity or alkalinity(neutral).       If you wish to check whether a chemical solution is acidic or basic, you use the instrument known as the *pH metre*         NOTE THIS...     The pH values range from *0 to 14*         This means that if you solve a problem under pH calculation and your answer as pH gives you a negative number or a number greater that 14, then your answer is likely to be wrong.       The lowest answer you should get in pH calculations is zero (0) while the highest is 14.     πŸ‘‡πŸΌπŸ‘‡πŸΌπŸ‘‡πŸΌπŸ‘‡πŸΌ *Mathematically, the pH of a solution is calculated using the formula*..πŸ‘‡πŸΌ       pH = – log[H^+]     Where [H^+] is the hydrogen ion concentration of the solution given to you or you call it the molarity if the solution.     Note that this is log to base 10.   *You need to know this*...πŸ‘‡πŸΌ       from the knowledge of mathematics, we can rearrange the pH formula as followsπŸ‘‡πŸΌ SinceπŸ‘‡πŸΌπŸ‘‡πŸΌ   pH = – log [H^+]   ThisπŸ‘†πŸ½is the same as πŸ‘‡πŸΌ     pH = –1 x log [H^+]   Remember the rule of logarithm which allows us to carry the –1 to the front as the power like thisπŸ‘‡πŸΌ     pH = log [H^+]^–¹   Also, according to the law of indices, any number raised to the power if –1 is the same as the inverse of that number (Ex. Y^–¹ = 1/Y)     Therefore,πŸ‘‡πŸΌ   *pH = log 1/[H^+]* βœ…     Note thisπŸ‘‡πŸΌπŸ‘‡πŸΌπŸ‘‡πŸΌ   pH = – log [H^+]   While   pOH = – log [OH^–]     For acids, we work with pH formula, but for bases, we work with pOH formula  
*If you are asked to find the pH of a substance, before you start solving the problem, check whether the compound given to you in the question is an acid or a base*       Do not rush into solving with the pH formula. Whether to use the first formulaπŸ‘‡πŸΌ pH = – log [H^+] Or thisπŸ‘‡πŸΌ pOH = – log [OH^–] depends on the compound given in the question     *If you are asked to find the pH of a substance, before you start solving the problem, check whether the compound given to you in the question is an acid or a base* Do not rush into solving with the pH formula. Whether to use the first formulaπŸ‘‡πŸΌ pH = – log [H^+] Or thisπŸ‘‡πŸΌ pOH = – log [OH^–] depends on the compound given in the question     NOTE THIS...πŸ‘‡πŸΌ [H^+] means πŸ‘‰πŸΌ *hydrogen ion concentrationof the given solution* or you call it the *molarity or concentration of the solution given in the question* while [OH^–] means πŸ‘‰πŸΌ *Hydrodyl ion concentration* or you call it the *molarity or concentration of the base given* in the question.         It is also very important for you to know that before substituting the value of the concentration of the substance into the pH formula or pOH formula, πŸ‘‰πŸΌ *you need to multiply the the concentration of the acid solution with the number of ionizable hydrogen's present (i.e basicity) in the acid given in the question*     or you *multiply the concentration of the basic solution with the number of ionizable hydroxyl ions present in the base given in the question*         For instance, if the acid given in the question is H2SO4 and the concentration given is *0.015*, then the value of [H^+] you will use while calculation for the   pH will be πŸ‘‰πŸΌ 0.005 x 2 = 0.01 Which means that the pH value will beπŸ‘‡πŸΌ pH = –log [H^+] = –log(0.01) ThereforeπŸ‘‡πŸΌ pH = 2.     Also, πŸ‘‰πŸΌ if a Ca(OH)2 base is given, the hydroxyl ion concentration [OH^–] will be multiplied by 2 before finding the pOH value because of the two ionizable OH groups present in the base     The possible answers you are expected to get when calculating for pH or pOH lies withing the range of *0 to 14*     . NOTE...πŸ‘‡πŸΌ pH answers can be a decimal number. If you get a negative value as your answer, you should know that something is wrong with your answer. πŸ‘‰πŸΌ Your pH or pOH answer should not be greater than 14 and should not be less than 0 (negative).     *There is this trick about questions on pH calculations you need to understand*               Here is the trick.. πŸ‘‡πŸΌπŸ‘‡πŸΌπŸ‘‡πŸΌ At times, you will be given a base in the question, and you will be asked to calculate the pH of the compound. *One important thing you need to know is this...*πŸ‘‡πŸΌ When the compound given in the question is an acid, and you are asked to find pH, you are free to use the general pH formula that says that..πŸ‘‡πŸΌ *pH = –log [H^+]* And the answer you get here becomes the answer.     πŸ‘‰πŸΌ *But, if the compound given in the question is a base, and you are asked to find the pH, do not use the pH formula since the compound given is a base*. What you should use first is the pOH formula which states that...πŸ‘‡πŸΌ *pOH = –log[OH^–]* When you use this formula and get an answer, the answer you got is the *pOH value* of the base given in the question.       Then, in order to find the pH value which you are looking for, you apply this formula below which states that *the sum of the pH and the pOH values of a substance is equal to 14*         ....πŸ‘‡πŸΌ *pH + pOH = 14* With thisπŸ‘†πŸ½, since you have found the pOH value, you can noe find the pH value which you are looking for. *Please note that you are using this method the substance given to you is a base and you are asked to find pH* But if the substance given in the question is an acid, and you are asked to find the pH, just use the pH formula straight and find your answer.     This same thing happens when the question gives you an acid and ask you to find pOH. In that case, you will have to find the pH of the acid first using the pH formula. After finding the pH, you now bring that same formula that says...πŸ‘‡πŸΌ     *pH + pOH = 14* from hereπŸ‘†πŸ½ you can now find the pOH.     But if the substance given in the question is a base and you are asked to find the pOH, you will use the pOH formula straight to find your answer.         *In summary,*...πŸ‘‡πŸΌ pH = –log[H^+] is used when an acid is given. while *pOH = –log[OH^–]* is used when a base is given. But after using this to get an answer, know that what you have gotten is *pOH value* and not *pH value*. To find pH value, you then applyπŸ‘‡πŸΌ *pH + pOH = 14.*   NOTE...πŸ‘‡πŸΌ πŸ‘‰πŸΌ *the product of the hydrogen ion concentration and hydroxyl ion concentration is equal to the ionization constant of pure water which is a constant and it's equal to 1.0x10–¹⁴     Mathematically,...πŸ‘‡πŸΌ     *[H^+] x [OH^–] = Kw* where Kw πŸ‘‰πŸΌ ionization constant of pure water. And Kw = 1.0x10–¹⁴   Now let us introduce other formulae in this topic     Also πŸ‘‰πŸΌ *The product of the dissociation constant of acid and the dissociation constant of base is equal to the ionization constant of pure water*       Mathematically, πŸ‘‡πŸΌ Ka x Kb = Kw Where Ka πŸ‘‰πŸΌdissociation constant of acid Kb πŸ‘‰ dissociation constant of base Kw πŸ‘‰πŸΌ ionization constant of pure water = 1.0x10^–¹⁴     Also..   When the concentration and *Ka* of a weak acid or base are given use the formula...πŸ‘‡πŸΌ pH = –log *(square root of C x Ka)* Where C = concentration Also for a weak base whose concentration and *Kb* are given,..πŸ‘‡πŸΌ pOH = –log *(square root of C x Kb)* where C = concentration       *In summary,*...πŸ‘‡πŸΌ pH = –log[H^+] is used when an acid is given. while *pOH = –log[OH^–]* is used when a base is given. But after using this to get an answer, know that what you have gotten is *pOH value* and not *pH value*. To find pH value, you then applyπŸ‘‡πŸΌ *pH + pOH = 14.*     NOTE...πŸ‘‡πŸΌ πŸ‘‰πŸΌ *the product of the hydrogen ion concentration and hydroxyl ion concentration is equal to the ionization constant of pure water which is a constant and it's equal to 1.0x10–¹⁴       Mathematically,...πŸ‘‡πŸΌ *[H^+] x [OH^–] = Kw*     where Kw πŸ‘‰πŸΌ ionization constant of pure water. And Kw = 1.0x10–¹⁴   Now let us introduce other formulae in this topic.     Also     πŸ‘‰πŸΌ *The product of the dissociation constant of acid and the dissociation constant of base is equal to the ionization constant of pure water* Mathematically, πŸ‘‡πŸΌ Ka x Kb = Kw Where Ka πŸ‘‰πŸΌdissociation constant of acid Kb πŸ‘‰ dissociation constant of base Kw πŸ‘‰πŸΌ ionization constant of pure water = 1.0x10^–¹⁴       Also...πŸ‘‡πŸΌ When the concentration and *Ka* of a weak acid or base are given use the formula...πŸ‘‡πŸΌ pH = –log *(square root of C x Ka)* Where C = concentration Also for a weak base whose concentration and *Kb* are given,..πŸ‘‡πŸΌ pOH = –log *(square root of C x Kb)* where C = concentration     Finally, let's talk about a *BUFFER SOLUTION*     As you know, the pH of a pure water is *7*.     This implies that pure water is *neutral*(not acidic and not basic). You know that if you drop some acid into water, the water will become acidic and the pH of the water will become less than 7.           Also If you drop a base into water, the water will become basic and the pH of the water will become greater than 7.     This is applicable to other chemical substances except for BUFFER SOLUTIONS.     A πŸ‘‰πŸΌ *BUFFER SOLUTION* is a solution whose pH value remains the same when a little drop of an acid or alkali is dropped into it. Or we can say thatπŸ‘‡πŸΌ   A *BUFFER SOLUTION* is a solution that resists a change in pH when a little drop of acid or base is dropped into it.     So you see, from the definition, I believe you understand when to call a substance a buffer solution.     If you drop an acid or a base into an unknown solution, and the pH value of that solution remains unchanged, then that solution is called a πŸ‘‰πŸΌ *Buffer solution*             *For a Buffer solution,* πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ» *pH = pKa + log[salt]/[acid]* πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ» Where pKa = –logKa [salt] = concentration of salt [acid] = concentration of acid   Also For a Buffer solution, *pOH = pKb + log[salt]/[base]* πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ» Where pKb = –logKb [salt] = concentration of salt [base] = concentration of base     Remember thatπŸ‘‡πŸ» Ka πŸ‘‰πŸΌ ionization constant of acid And Kb πŸ‘‰πŸΌ ionization constant of base     Before using the BUFFER SOLUTION pH or pOH formula, you must find the following first πŸ‘‡πŸ»         FOR pH πŸ‘‰πŸΌ –log Ka πŸ‘‰πŸΌ concentration of salt πŸ‘‰πŸΌ concentration of acid       FOR pOH πŸ‘‰πŸΌ –log Kb πŸ‘‰πŸΌ salt concentration πŸ‘‰πŸΌ acid concentration     NOTE... πŸ‘‡πŸ»       Sometimes, you may not be given the acid, base or salt concentrations directly.     In that case, you will be given the variables you need to use and find those concentrations before you can then fine the pH or pOH you are looking for.       You must always remember thatπŸ‘‡πŸ» C = n/v WhereπŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ» C = concentration n = mole v = volume in dmΒ³ or in litre. Also remember that Mole(n) = mass/molar mass     πŸ‘‰πŸΌ *FOR pH* πŸ‘‰πŸΌ pH = –log[H^+]     πŸ‘‰πŸΌ pH –log(square root of *Ka x C*     Where C = acid concentration       For Buffer solution,.. πŸ‘‡πŸ» πŸ‘‰πŸΌ pH = pKa + log[salt]/[acid]     πŸ‘‰πŸΌ *FOR pOH* πŸ‘‰πŸΌ pH = –log[OH^-] πŸ‘‰πŸΌ pH –log(square root of *Kb x C*     Where C = base concentration For Buffer solution,.. πŸ‘‡πŸ» πŸ‘‰πŸΌ pOH = pKb + log[salt]/[base]     *Then we have these general formulae* πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ»πŸ‘‡πŸ» pH + pOH = 14     πŸ‘‰πŸΌ *[H^+] x [OH^-] = 1.0 x 10^-14*     πŸ‘‰πŸΌ *Ka x Kb = Kw*     Where Kw = ionization constant of water (Kw = 1.0x10^-14)       *QUESTION ONE* Calculate the pH of a solution of calcium hydroxide whose molarity is 0.0005.           *QUESTION TWO* Given that the molar concentration of a tribasic acid is 0.003mol/dmΒ³. Calculate the pOH value of the acid.         *QUESTION THREE* An unknown solution contains 0.086moles of of hydrogen ions in 500cmΒ³. The solution is likely to be ___? (A). Basic (B). Acidic   (C). Neutral (D). Buffer       You can now go back to the site and solve the calculation questionsΒ  on pH and pOH calculations.     GOODLUCK.
Subscribe
Notify of
guest
0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x