Calculate the pH of each of the following strong acid solutions. (a) 0.00704 M HIO4 pH = (b) 0.696 g of HIO4 in 25.0 L of solution pH = X PH = (c) 70.0 mL of 8.00 M HIO4 diluted to 3.40 L X pH = X (d) a mixture formed by adding 16.0 mL of 0.00197 M HIO4 to 16.0 mL of 0.00284 M HIO4 X

6 4 The active ingredient formed by aspirin in the body is salicylic acid, CH₂OH(CO₂H). The carboxyl group (-CO,H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. What are all of the ionic and molecular species present in a 0.001-M aqueous solution of CH₂OH(CO₂H) in order of descending concentration? 4 A B C D [H₂O]> [C_H_OH(CO₂H)]> [H*]>[C_H₂OH(CO₂) -]>[C_H₂O(CO₂H¯)]> [OH-] [H₂O]>[H+]>[C_H₂OH(CO₂H)]>[C_H_OH(CO₂)¯]>[C.H₂O(CO₂H¯)]>[OH-] [H₂O]> [C_H_OH(CO₂H)]> [CH₂OH(CO₂)-]> [C_H_O(CO₂H-)]> [OH-]>[H+] [+H]] ] <[0²H]

What is the pH of a solution with molar concentration of H*, [H*] = 1.0 x 10-9 M?

Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H2C2O4) in a solid given that a 0.7984-g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

Chemists use a number denoted by pH to describe quantitatively the acidity or basicity of solutions. Bydefinition, pH = - log [H+], where [H+] is the hydrogen ion concentration in moles per liter.

Calculate pH of the following solutions: a) 0.25 M HNO3 in 0.065 M CH3COOH (K,=1.75×10). b) 0.33 M NH,NO, (K, for NH4=5.6×10 10 ; K, for HNO2=4.0×10“) c) 0.33 M NHẠNO3 (K, for NH4=5.6x10 10) d) 9.95x1011 M Ba(OH)2

The pH of an acidic solution is 3.37. What is [H*1? 80 F1 F2 F3 F4 F6 F7 F8 @ # $ % & 2 3 4 6 7 8 9 W E R T Y 云 5

Calculate pH and pOH for each of the following solutions and determine whether thesolution is acidic, basic or neutral:(i) [H3O+] = 4.1 x 10-10 M (ii) [OH–] = 7.6 x 10-3 M(iii) [OH–] = 6.9 x 10-4 M (iv) [H3O+] = 3.5 x 10-6 M

Human blood has pH=7.4 .  Therefore, its (a) [OH-] =3.28x10-8(M)    (b) [H+] =3.28x10-8(M)     (c) [H+] =3.28x10-7(M)     (d) [H+] = 10-7.4  (M)    (e) pOH=8.6

1. (EtO)₂C=O, NaOEt 2. HCI