Question

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

(a) 0.00293 M HI

pH =

(b) 0.899 g of HIO4 in 50.0 L of solution

pH =

(c) 61.0 mL of 3.80 M HI diluted to 1.50 L

pH =

(d) a mixture formed by adding 34.0 mL of 0.00899 M HI to 75.0 mL of 0.00271 M HIO4

pH =

Answer 1

(a) given 0.00293 M HI

[H⁺] = 0.00293 M

p^H = - log[H⁺]

= - log(0.00293)

pH = 2.53

(b) molarity = weight/ (gram molecular weight * volume in L)

= 0.899 / (191.9 * 50)  

= 9.369 X 10^-5

p^H = - log[H⁺]

= - log( 9.369 X 10^-5 )

pH = 4.028

(c)

M = 3.80 X 61 x 10^-3 / 1.50 = 0.154 M

p^H = - log[H⁺]

= - log( 0.154 )

pH = 0.812

(d) total moles = 34 X 0.00899 + 75 X 0.00271 = 0.5089 moles

molarity = 0.5089 / (34 + 75)

= 4.67 X 10^-3 M

p^H = - log[H⁺]

= - log( 4.67 X 10^-3 )

pH = 2.331

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