229829 20 mL of 0.1M acetic acid is mixed in a solution of NaOH. If 10 mL of 0.1MNaOH is present then H+concentration in resulting solution is (Ka of acetic acid =1.7×10−5 )
Given, Ka=1.7×10−3Solution of NaOH=20mlCH3COOH+NaOH→CH3COONa+H2O20×0.110×0.12M mole 1M mole 1M mole 01M mole We know that,pOH=pKb+log salt acid Therefore,pH=pKa(H+)=Ka=1.7×10−5M
229830 Which of the following have maximum pH ?
Exp:{cc}| Material | $$ ||---|---||i. Black coffee | -5.0 ||ii. Blood | -7.4 ||iii. Gastric Juicev | −1.8−2.0 ||iv. Saliva | -6.8 ||
229831 Ksp of M(OH)2 is 3.2×10−11. The pH of saturated solution in water is
M(OH)2→M2++2OH−Ksp=[x][2x]2]=4x33.2×10−11=4x30.8×10−11=x3x3=8×10−12x=2×10−4[OH]=4×10−4MpOH=−log[OH]=4−2log2=4−2×0.03=3.4We know,pH+pOH=14pH=14−3.4=10.60
229835 40 mL of 0.1M ammonia solution is mixed with 40 mL of 0.1MHCl. What is the pH of the mixture? (pK pKb of ammonia solution is 4.74).
Henderson - Haselbalch equationpOH=pKb+log( salt base )pOH=pKb+log(NH4ClNH3)At half stage of titration salt = Base as pOH=pKb=4.74So, pH=14−pOH=14−4.74=9.26
229836 When 10 mL of 0.1M acetic acid (pKa=5.0), is titrated against 10 mL of 0.1M ammonia solution (pKb=5.0), the equivalence point occurs at pH :
CH3COOH+NH4OH→CH3COONH4+H2OHere, given thatpKa=5pKb=5pKa=−logKa and pKb=−logKbpH=−12[logKa+logKW−logKb]=−12[−5+log10−14−(−5)]=−12[−5−14+5]=−12×(−14)=7Thus, the end point or equivalence point is obtained at pH=7 in neutral medium.
