pH of #0.1(M)# #NaH_2PO_4# is what?

2 Answers
Aug 8, 2017

Answer:

pH = 4.10

Explanation:

If it was a strong acid then the concentration of #H^+# that dissociates from #NaH_2PO_4# would be #2 xx 0.1 = 0.2M#

But #NaH_2PO_4# is a weak acid. It will dissociated partially.

If #x# represents concentration of acid that dissociates then

#x = sqrt(K_a xx C)#
(see Ernest answer for more details about this formula)

In order to find pH of a weak acid we should know acid dissociation constant (#K_a#) value.

#K_a# of #NaH_2PO_4# is #6.2 xx 10^-8#

#x = sqrt(6.2 xx 10^-8 xx 0.1)#

#x = 7.9 xx 10^-5 M#

#pH = -log[H^+]#

#pH = -log[7.9 xx 10^-5]#

#pH = 4.10#

Aug 9, 2017

Answer:

#"pH = 4.10"#

Explanation:

#"NaH"_2"PO"_4# dissociates completely in solution:

#"NaH"_2"PO"_4 → "Na"^"+" + "H"_2"PO"_4^"-"#

The dihydrogen phosphate ion is a weak acid:

#"H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; K_text(a) = 6.23 × 10^"-8"#

We can use an ICE table to calculate the concentrations of the ions in solution.

Let's rewrite the equation as

#color(white)(mmmmmmmm)"HA"^"-" +color(white)(ll) "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"2-"#
#"I/mol·L"^"-1":color(white)(mmm)0.1color(white)(mmmmmmmll)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1":color(white)(mm)"0.1 -"xcolor(white)(mmmmmmm)xcolor(white)(mmm)x#

#K_text(a) = (["H"_3"O"^"+"]["A"^"2-"])/(["HA"]) = (x × x)/("0.1 -"color(white)(l)x) = x^2/("0.1 -"color(white)(l)x) = 6.23 × 10^"-8"#

Check for negligibility:

#0.1/(6.23 × 10^"-8") = 2 × 10^6 ≫ 400#.

#x ≪ 0.1#.

Then

#x^2/0.1 = 6.23 × 10^"-8"#

#x^2 = 0.1 × 6.23 × 10^"-8" = 6.2 × 10^"-9"#

#x = 7.9 × 10^"-5"#

#["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 7.9 × 10^"-5"color(white)(l)"mol/L"#

#"pH" = "-log"["H"_3"O"^"+"] = "-log"(7.9 × 10^"-5") = 4.10#