The chemical equation is
#"HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+"#
We can use an ICE table to solve the problem.
#color(white)(mmmmmll)"HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+"#
#"I/mol·L"^"-1": color(white)(mll)1.65 color(white)(mmmmmmmml)0color(white)(mmmm)0#
#"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmmmml)"+"xcolor(white)(mmll)"+"x#
#"E/mol·L"^"-1": color(white)(m)"1.65-"xcolor(white)(mmmmmmmm)xcolor(white)(mmmll)x#
#K_text(a) = (["HCOO"^"-"]["H"_3"O"^"+"])/(["HCOOH"]) = x^2/(1.65-x) = 1.8 × 10^"-4"#
Check for negligibility:
#1.65/(1.8 ×10^"-4") = 9200 > 400#. ∴ x ≪ 1.65.
Then
#x^2/1.65 = 1.8 × 10^"-4"#
#x^2 = 1.65 × 1.8 × 10^"-4" = 2.97 × 10^"-4"#
#x = 0.0172#
#["H"_3"O"^"+"] = "0.0172 mol/L"#
#"pH = -log"["H"_3"O"^"+"] = "-log"0.0172 = 1.76#