Your general expression should be in .NET framework

Generate EAN-13 Supplement 5 in .NET framework Your general expression should be

Calculate the Ka for a 050M solution of benzoic acid, C6H5COOH, if the [H+] concentration is 56 10 3

If you have a 02M solution of ammonia, NH3, that has a Kb of 180 10 5, what is the pH of that solution

In this reaction, HCl donates a proton to H2O, making it the Br nsted-Lowry acid Water, which accepts the proton, is the Br nsted-Lowry base This makes H3O+ the conjugate acid and Cl the conjugate base Water can act as the base in this reaction and as an acid in the example problem because it s composed of both a hydrogen ion and a hydroxide ion and can therefore either accept or donate a proton For Arrhenius acids, remember to track the movement of H+ and OH ions If the reaction yields an OH product, the substance is a base, while an H+ product reveals the substance to be an acid a Arrhenius base NaOH dissociates in water to form OH ions, making it an Arrhenius base b Arrhenius acid HF dissociates in aqueous solution to form H+ ions, making it an Arrhenius acid

To identify Lewis acids and bases, track the movement of electron pairs Draw a Lewis dot structure to locate the atom with a lone pair available to donate This is the Lewis base a H2O is the Lewis base, while Cr3+ is the Lewis acid Your Lewis dot structure should show that the lone pair of electrons is donated to the bond by H2O

b NH3 is the Lewis base, while Ag+ is the Lewis acid Your Lewis dot structure should show that the lone pair of electrons is donated to the bond by NH3

c The negative charge on the chlorine indicates that it s electron rich, making it the Lewis base HgCl2 is, therefore, the Lewis acid

To calculate the pH values in this problem, use the equation pH = log [H+] and plug in all known values If you re given [OH ] instead of [H+], simply calculate the pOH (using the equation pOH = log[OH ]) instead and then subtract that number from 14 to give you the pH a pH = 13 Because the coefficient on the H+ concentration is 1, the pH is simply the exponent b pH = 88 Plug the given [H+] into the pH equation, giving pH = log (158 10 9) = 88 c pH = 73 First, use the OH concentration to calculate the pOH pOH = log(2 10 7) = 67 Then use the relationship between pH and pOH (14 pOH = pH) to determine the pH: 14 67 = 73 d pH = 7 Because the coefficient of this OH concentration is 1, you can read the pOH directly from the exponent to get pOH = 7 Then use the relationship between pH and pOH (14 pOH = pH) to determine the pH: 14 7 = 7

To calculate pOH, simply take the negative logarithm of the OH concentration If you re given the H+ concentration, use it to calculate the pH and then subtract that value from 14 to yield the pOH a pOH = 63 First, use the H+ concentration to calculate the pH pH = log(2 10 8) = 77 Then use the relationship between pH and pOH (14 pH = pOH) to determine the pOH: 14 77 = 63 b pOH = 103 Here, you can simply use the OH concentration to calculate the pOH directly pOH = log (51 10 11) = 103

To determine acidity, you need to calculate pH a Basic Use [OH ] to determine pOH pOH = log(25 10 7) = 66 Subtract this value from 14 to get a pH of 74 This solution is very slightly greater than 7 and, therefore, slightly basic b Acidic Use [OH ] to determine pOH pOH = log(31 10 13) = 125 Subtract this value from 14 to get a pH of 15 This solution is significantly smaller than 7 and, therefore, quite acidic c Acidic Use the [H+] concentration to determine pH pH = log(421 10 5) = 44 d Basic Use the [H+] concentration to determine pH pH = log(89 10 10) = 91

In this problem, use the formula [H+] = 10 pH If you re given the pOH instead of the pH, begin by subtracting the pOH from 14 to give you the pH and then plug that number into the formula a 50 10 4 Use the pH to calculate the H+ equation using the equation [H+] = 10 pH [H+] = 10 33 = 50 10 4 b 204 10 8 Use the pH to calculate the H+ equation using the equation [H+] = 10 pH [H+] = 10 769 = 204 10 8 c 162 10 4 Begin by using the pOH to calculate the pH using the equation 14 pOH = pH 14 1021 = 379 = pH Then use the pH to calculate the H+ equation using the equation [H+] = 10 pH [H+] = 10 379 = 162 10 4 d 182 10 13 Begin by using the pOH to calculate the pH using the equation 14 pOH = pH 14 126 = 1274 = pH Then use the pH to calculate the H+ equation using the equation [H+] = 10 pH [H+] = 10 1274 = 182 10 13