Question #6f85f

Question

2807 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-07T19:58:33

Here's what I get.

a)

$["OH"^"-"] = 1.83 × 10^"-7"color(white)(l) "mol/L"$

$["H"_3"O"^"+"] = (1.00 × 10^"-14")/(1.83 × 10^"-7") color(white)(l)"mol/L" = color(white)(l)5.46 × 10^"-8" color(white)(l)"mol/L"$

b)

$["H"_3"O"^"+"] = 8.15 × 10^"-9"color(white)(l) "mol/L"$

$["OH"^"-"] = (1.00 × 10^"-14")/(8.15 × 10^"-9") color(white)(l)"mol/L" = color(white)(l)1.23 × 10^"-6" color(white)(l)"mol/L"$

c)

$"H"_3"O"^"+"] = "0.000 665" color(white)(l) "mol/L"$

$["OH"^"-"] = (1.00 × 10^"-14")/("0.000 665") color(white)(l)"mol/L" = color(white)(l)1.50 × 10^"-11" color(white)(l)"mol/L"$

a) and b) are basic because, in each, $["H"_3"O"^"+"] < 1.00 × 10^"-7" color(white)(l)"mol/L"$