Question

Calculate the pH of each of the following strong acid solutions.

(a) 0.00471 M HClO4 pH =

(b) 0.329 g of HBr in 16.0 L of solution pH =

(c) 65.0 mL of 7.50 M HClO4 diluted to 4.60 L pH =

(d) a mixture formed by adding 54.0 mL of 0.00155 M HClO4 to 47.0 mL of 0.00191 M HBr pH =

Answer 1

(a) HClO4 ===> H+ + ClO4

So, pH = -log[H+] = -log [0.00471] = 2.32

(b) no. of moles = mass taken/molar mass = 0.329/80.91 = 4.06E-3moles

molarity = no. of moles/volume = 4.06E-3moles/ 16L = 2.54E-4M

pH = -log[H+] = -log [2.54E-4M] = 3.59

(c) M1V1 = M2V2

7.5M x 0.065L = M2 x 4.60L

M2 = 0.106M

So, pH = -log[H+] = -log [0.106] = 0.97

(d)no. of Moles of HBr present = 0.00191M x 0.047L = 8.977E-5 mol

no. of Moles of HClO4 = 0.00155M x 0.054 = 8.37E-5moles

Total moles = 8.977E-5 mol + 8.37E-5moles = 1.7347E-4moles

molarity = 1.7347E-4moles/ (0.054+ 0.047) = 1.71E-3M

pH = - log (1.71E-3M) = 2.76