Find the #["H"_3"O"^(+)]# and #"pH"# for the following weak acids or bases?

(a) #["H"_3"O"^"+"] = 9.95 × 10^"-6"color(white)(l) "mol/L"; "pH = 5.00"#
#color(white)(ml)["H"_3"O"^"+"] = 1.00 × 10^"-6"color(white)(l) "mol/L"; "pH = 5.00"#

(b) #["H"_3"O"^"+"] = 7.07 × 10^"-6"color(white)(l) "mol/L"; "pH = 5.15"#
#color(white)(ml)["H"_3"O"^"+"] = 1.00 × 10^"-5"color(white)(l) "mol/L"; "pH = 5.00"#

(c) #["H"_3"O"^"+"] = 1.64 × 10^"-5"color(white)(l) "mol/L"; "pH = 4.79"#
#color(white)(ml)["H"_3"O"^"+"] = 2.16 × 10^"-5"color(white)(l) "mol/L"; "pH = 4.66"#

(d) #["H"_3"O"^"+"] = 4.59 × 10^"-9"color(white)(l) "mol/L"; "pH = 8.54"#
#color(white)(ml)["H"_3"O"^"+"] = 4.54 × 10^"-9"color(white)(l) "mol/L"; "pH = 8.54"#

(f) #["H"_3"O"^"+"] = 8.26 × 10^"-8"color(white)(l) "mol/L"; "pH = 7.08"#
#color(white)(ml)["H"_3"O"^"+"] = 8.26 × 10^"-8"color(white)(l) "mol/L"; "pH = 7.08"#

(a) 0.10 mol/L #"NaHB"#

The equilibria are

#"H"_2"B" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HB"^"-"; "p"K_text(a1) = 3.0#
#"HB"^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "B"^"2-"; color(white)(ll)"p"K_text(a2) = 7.0#

#K_text(a1) = 10^"-3.0" = 1.00 × 10^"-3"#
#K_text(a2) = 10^"-7.0" = 1.00 × 10^"-7"#

The exact formula for calculating #["H"_3"O"^"+"]# is

#["H"_3"O"^"+"] = sqrt((K_text(a1)K_text(a2)["HB"^"-"] + K_text(a1)K_text(w))/(["HB"^"-"] + K_text(a1)#

#["H"_3"O"^"+"] = sqrt((1.00 × 10^"-3" × 1.00 ×10^"-7" × 0.10 + 1.00 × 10^"-3"× 1.00 × 10^"-14")/(0.10 + 1.00 × 10^"-3")) = sqrt((1.00 ×10^"-11" + 1.00 × 10^"-17")/0.101) = sqrt((1.00× 10^"-11")/0.101) = sqrt(9.90 × 10^"-11")#
#= 9.95 ×10^"-6"#

#"pH = -log"["H"_3"O"^"+"] = "-log"(9.95 × 10^"-6") = 5.00#

The approximate formula for calculating #["H"_3"O"^"+"]# is

#["H"_3"O"^"+"] = sqrt(K_text(a1)K_text(a2))#

#["H"_3"O"^"+"] = sqrt(1.00 × 10^"-3" × 1.00 × 10^"-7") = sqrt(1.00 ×10^"-10") = 1.00 ×10^"-5"#

#"pH = -log"(1.00 × 10^"-5") = 5.00#

Rather than do the rest of the calculations manualy, I will let Microsoft Excel do them.

Here are my results for Part (a).

(b) 0.0010 mol/L #["HB"^"-"]#

(c) 0.010 mol/L #"NaH"_2"PO"_4#

The equations are

#"H"_3"PO"_4 + "H"_2"O" ⇌ "H"_3"O"^"+" + "H"_2"PO"_4"-"; K_text(a1) = 7.52 ×10^"-3"#
#"H"_2"PO"_4"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; color(white)(l)K_text(a2) = 6.23 × 10^"-8"#

(d) 0.010 mol/L #"NaHCO"_3#

The equations are

#"H"_2"CO"_3 + "H"_2"O" ⇌ "H"_3"O"^"+" + "HCO"_3^"-"; K_text(a1) =4.3 × 10^"-7" #
#"HCO"_3^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "CO"_3^"2-"; color(white)(m)K_text(a2) = 4.8 × 10^"-11" #

(e) No data

(f) 0.010 mol/L #"NaSH"#

The equations are

#"H"_2"S" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HS"^"-"; K_text(a1) = 1.1 × 10^"-7"#
#"HS"^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "S"^"2-"; color(white)(ll)K_text(a2) = 6.2 × 10^"-8"#