H2SO4 몰농도. 9.72 mL H2SO4 react 0.335 g NaOH

What is the molar concentration of a solution of sulfuric acid, H2SO4,

if 9.72 mL react completely with 0.335 g of NaOH

which was dissolved in enough water to make 200. mL of solution?

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NaOH의 몰질량 = 40.00 g/mol 이므로,

0.335 g / (40.00 g/mol) = 0.008375 mol NaOH

H2SO4 + 2NaOH → Na2SO4 + 2H2O

H2SO4 : NaOH = 1 : 2 계수비(= 몰수비) 이므로,

0.008375 mol NaOH와 반응하는 H2SO4의 몰수를 계산하면,

H2SO4 : NaOH = 1 : 2 = ? mol : 0.008375 mol

? = 0.008375 / 2 = 0.0041875 mol H2SO4

H2SO4 용액의 몰농도를 계산하면,

몰농도 = 용질 mol수 / 용액 L수

( 참고 https://ywpop.tistory.com/3222 )

= 0.0041875 mol / (9.72/1000 L)

= 0.4308 mol/L

답: 0.431 M

[키워드] H2SO4 + NaOH 중화 기준, H2SO4 + NaOH 기준