Which structure is the sun located

Explanation:

C₈H₆Cl₂O₃

mol weight

= 8 x 12 + 6 x 1 + 35.5 x 2 + 3 x 16

96 + 6 + 71 + 48 = 221 g

230 g = 230 / 221 mol = 1.04 M solution

pKa = 2.73

Ka = 10⁻²°⁷³

= 1.86 x 10⁻³

Let the compound be represented by KH

KH = K⁻ + H⁺

a        0       0

a - x      x        x  

x ² / (a - x)  = Ka

x² / (1.04 - x) = .00186

x is very small so 1.04 - x = 1.04

x² / 1.04 = .00186

x² = .0019344

x = .044

= 44 x 10⁻³

[ H⁺ ] = 44 x 10⁻³

pH = - log ( 44 x 10⁻³)

= 3 - log 44

= 1.35