Answer:
Given info : 500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. To find : will the resulting solution is acidic , basic or neutral ? Calculate the molarity of the dilute solution. solution : no of moles of Na2CO3 = normality/n %3D - factor x volume 2/2 x 500/1000 = 0.5 mol %D no of moles of H2SO4 = 3/2 x 400/100O = 0.6 mol %3D We see, Na2CO3 + H2S04 => Na2S04 + CO2 + H2O Here one mol of Na2C03 reacts with one mole of H2SO4. So, 0.5 mol of Na2CO3 reacts with 0.5 mol of H2SO4. so, remaining 0.1 mol of H2SO4 makes solution acidic. Now molarity of solution = remaining no of moles of H2SO4/volume of solution= 0.1/1 = %3D 0.1M
Answer:
7.58
Explanation:
Given that
A buffer solution obtained by dissolving 13.0 g of KH₂PO₄ and 26.0 g of Na₂HPO₄ in water and then diluting to 1.00 L.
pKa = 7.21.
To get the PH we are going to use Henderson - Hasselblach equation:
PH = Pka + ㏒ [A/AH]
when the molar mass of Na2HPO4 = 142 g/mol
and A is the conjugate base HPO4-- ions so,
∴[A] = 32g / 142 g/mol
= 0.225 M
and when the molar mass of KH2PO4 = 136 g/mol
and AH is the weak acid H2PO4- ions so,
∴[AH] = 13 g / 136 g/mol
= 0.096 M
and when we have the Pka value of H3PO4 = 7.21
so, by substitution:
∴ PH = 7.21 + ㏒ (0.225 / 0.096)
= 7.58
35.453 (chlorine) + 10.811 (boron) = 46.264