With mathematics in mind you can write such a simple equation with variables a, b, c denoting the coefficients to be found:Īll you have to do is balance the quantity for each element on the left and the right side of the equation. Use the calculator with the explanation box checked to see the output. To see how the program finds the solution let`s start from a simple example.Īll the examples can be cut and pasted into the calculator That students will find it very useful in determining the To prove that this algebraic technique works. It was a real shock when I realized the power of It provides better solutions than chemistry itself! SuchĪn idea might seem a bit strange for a chemist. Mathematics provides a general way to find reaction coefficients. Luckily it turns out that oxidation numbers and half-reaction are unnecessary! Another method exists that uses algebra to find the In some cases however it is hard to use this method because oxidation numbers cannot be unequivocally attributed. The classical approach uses oxidation numbers and many chemists cannot imagine any way to find the solution other than balancing half-reactions for the process.
Remember that = :ĥ) The K b of X¯ is related to the K a of HX (what we want to get) by way of K w.Finding coefficients for chemical reactions is often complicated, especially in the case of redox processes. = 10¯ pOH = 10¯ 5.10 = 7.9433 x 10¯ 6 M (keep some guard digits)Ĥ) Now, we can write the K b expression. The pH is where we start:ģ) We use the pOH to give us the two values in the numerator of the K b expression for X¯ Based on the fact that HX is the conjugate base, we will use K aK b = K w to get the K a.Ģ) We need to fill in the right side of the K b expression for X¯. Solution technique: we will determine the K b of X¯ (since that is what we have data for). K b = K w / K a = 1.00 x 10¯ 14 / 4.5 x 10¯ 4 = 2.2222 x 10¯ 11īonus Example: Determine the K a of the weak acid HX knowing that a 0.10 M solution of LiX has pH = 8.90. (Use the equation K w = K aK b to go from the K a of the acid to the K b of its conjugate base.)ġ) We need to get the K b of the propionate ion first:Ģ) Now, the solution follows the pattern outlined in the tutorial:Įxample #5: Given that the K a for HOCl is 3.5 x 10¯ 8, calculate the pH of a 0.102 M solution of Ca(OCl) 2ġ) A 0.102 M solution of Ca(OCl) 2 is 0.204 M in just OCl¯ ions (which are the conjugate base of the acid HOCl)ģ) Conjugate pairs have the following relationship:Ĥ) For the chemical equilibrium given in step 2, we have:Įxample #6: Calculate the pH of a 0.18 M solution of the weak base NO 2¯. The K b for C 6H 5COO¯ (benzoate ion) is 1.55 x 10¯ 10.Įxample #4: Find the pH of a 0.20 M solution of sodium propionate (C 2H 5COONa), where the K a of propionic acid = 1.34 x 10¯ 5. Since this is a base calculation, we need to do the pOH first:Įxample #3: Find the pH of a 0.30 M solution of sodium benzoate, C 6H 5COONa. X = 1.025 x 10¯ 3 M <- this is the ĥ) We then calculate the pH. If it turns out that the approximation is invalid, then a quadratic equation must be used to solve the problem.Įxample #2: What is the pH of a 0.0500 M solution of KCN? K b = 2.1 x 10¯ 5.ġ) Here is the chemical reaction (net ionic) for the hydrolysis of KCN:Ĥ) Ignoring the minus x in the usual manner, we proceed to sove for the hydroxide ion concentration: In other words, you will find that the initial concentrations are large enough and the K b are small enough that the 5% rule is a valid approximation. You may verify on your own that in each case, the following is true: In all the remaining examples we will not repeat the 5% rule calculation. Since this is a base calculation, we need to do the pOH first: (I will use NaAc as shorthand)ġ) Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:ģ) We can then substitute values into the K b expression in the normal manner:Ĥ) If we ignore the minus x (that is, we assume the 5% approximation to be true), we proceed to solve for the hydroxide ion concentration:ĥ) We verify the assumption (which is that 0.100 − x ≈ 0.100) by seeing if the following ratio is less than 5%:Ħ) We then calculate the pH. You will see such a situation starting in the fourth example as well as scattered through the additional problems.Įxample #1: What is the pH of a 0.100 M solution of sodium acetate? K b = 5.65 x 10¯ 10. Often, these problems are given with the K a of the acid and you have to calculate the value of the K b. Note: in the first three problems, I give the K b of the conjugate base (for example, the acetate ion in Example #1). ChemTeam: Hydrolysis calculations: salts of weak acids are bases Hydrolysis calculations: salts of weak acids are bases Hydrolysis Problems #1 - 10 Calculations with salts of weak bases Calculation involving the salt of a weak acid and a weak base Hydrolysis Problems #11 - 20 Intro to Hydrolysis Calculations Acid Base Menu
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