A 75,0 mL sample of 0.0500 MHCN (K-6.2 x10-10) is titrated with 0.421 M NaOH to...

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Question

science chemistry

Answer 1

Answer #1

At equivalent point : M1V1 = M2V2

0.0500M× 75.0ml = 0.421M × V2

V₂ = 8.907ml

HCN + NaOH $\rightarrow$ NaCN + H₂O

[NaCN] = 0.0500M× 75.0ml/(75.0+8.907)ml = 0.0447M

pH = 7 + 1/2(pKa + log c)

= 7 + 1/2(-log(6.2×10^-10) + log(0.0447)

= 10.9289

pH = 10.9

Answer : (c) 10.9

Answer 2

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