A solution is prepared to be 0.1475 M in Sr(OH)2. How do you find the [H3O+], [OH‐], pH, and pOH?

1 Answer
Jun 10, 2016

Answer:

#pH = 13.029" and " pOH = 0.9710#

Explanation:

Strontium Hydroxide ionizes in water to produce the strontium ion and the hydroxide ion. Note that #Sr(OH)_2# is slightly soluble in water.

# " "Sr(OH)_2 \ \ rightleftharpoons \ \ \ Sr^(2+) + \ \ \ 2OH^-#

#I " " 0.1475 \ M " " - " " -#

#C" " -x " "+x " "+ 2x#

#E" "0.1475-x " " x " " 2x#

#K_b=( [ OH^-]^2 [ Sr^(2+)])/([Sr(OH)_2]#

#K_b= ( (2x)^2*(x))/(0.1475-x)= 6.5xx10^-3#

Solve for #x#

# x = 0.05346 \ M#

# [ OH^-] = 2x= 2xx 0.05346 = 0.1069 M#

# -------------------#

In any aqueous solution, both ions #H_3O^+#and #OH^-# are present and they must satisfy the following condition:

#[H_3O^+] [ OH^-] = K_W#

#[H_3O^+] [ OH^-] = 1.0xx10^(-14)#

# -------------------#

#[OH^-] = 0.1069 \ M # ( determined)

#[H_3O^+] = (1.0xx10^-14)/([ OH^-] )#

#[H_3O^+] = (1.0xx10^-14)/(0.1069 )#

#[H_3O^+] = 9.355^-14 \ M#

#-------------------#

#pH =- log [H_3O^+]#

#pH =- log [\ 3.955^-14 \]#

#pH = 13.0290#

#pOH =- log [\ OH^-\]#

# pOH =- log [\ 0.1069\ ]#

#pOH = 0.9710#