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4 years ago

A solution with a pH of 11 has a [H+] of: 11 1 x 10-3 1 x 10-11 1 x 1011 1 x 103

Mathematics

1 answer:

serious [3.7K]4 years ago

4 0

A solution with a pH of 11 has a [H+] of $1.0\times 10^{-11}$

Option B is correct option.

Step-by-step explanation:

A solution with a pH of 11 has a [H+] of= ?

pH=?

Concentration of H+ [H+]= ?

Formula used: $pH=-log[H+]$

Putting values and finding [H+]

$pH=-log[H+]\\\,[H+]\,=10^{-pH}\\\,[H+]\,=10^{-11}\\\,[H+]\,=1.0\times 10^{-11}$

So, A solution with a pH of 11 has a [H+] of $1.0\times 10^{-11}$

Option B is correct option.

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The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

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so the base case holds. Assume the claim holds for $n=k$ , so that

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Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
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So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

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$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

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which agrees with the result obtained for part (c).

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