Question

#26. Calculate [OH ^-] and pH for each of the following solutions.

(a) 0.0057 M KOH

[OH-] = ___ M

pH =___

(b) 0.0476 g of KOH in 490.0 mL of solution

[OH -] =___ M

pH = ___

(c) 11.0 mL of 0.00474 M Sr(OH)₂ diluted to 700 mL

[OH -] = ___ M

pH =___

(d) A solution formed by mixing 13.0 mL of 0.000730 M Sr(OH)₂ with 27.0 mL of 1.3 x 10^-3 M KOH

[OH -] =___ M

pH = ___

Answer 1

a) 0.0057M KOH solution

[OH-] = 0.0057M = 5.7x 10^-3 M

pH = 14- pOH = 14 -( 3- log 5.7) = 11.7559

b) 0.0476 g of KOH in 490 mL

Molarity = [0.0476 x 1000]/ [ 46x490] =2.1x 10^-3 M

thus pH = 14 -( 3 - log2.1) = 11.3032

c) 11ml 0.00474M Sr (OH)2 diluted to 700mL

For dilution V1M1 - V2M2

New molariity = 11x4.74x 10^-3 /700= 7.44 x 10^-5 M

The [OH-] = 2 x molarity = 2x 7.44 x10^-5 = 1.488x 10^-4 M

pH = 14 - ( 4 -log 1.488) = 10.6884

d) 13ml 0.000730 Sr(OH)2 mixed with 27 ml 1.3 x10^{-3 M} KOH

[OH-] in sr (OH)2 = 13 x 7.3 x10^-4 = 9.49x10^-3

[OH-] on KOh = 27x 1.3x10^-3   =35.1x 10^-3

Thus new [OH-] in the 40mL (13+27) solution = [ 9.49x10^-3 +35.1x 10^-3] = 4.459 x 10^{-2 M}

hence pH = 14 - (2 - log 4.459) = 12.6493