Kinetics question regarding Rate Law The acid catalysed hydrolysis of an compound $\ce{A}$ at $\pu{303K}$:
$$\ce {A -&gt;[$k$] B}$$ has a half-life of $\pu{100 min}$ when carried out in a buffer solution of $\mathrm{pH=5}$ and $\pu{10 min}$ when carried out at $\mathrm{pH=4}$. Both the times the half-life are independent of the initial concentration of $\ce{A}$. Find its Rate Law.


I am planning something like this:
Let $$\text{Rate} = k [\ce{A}][\ce{H+}]^x\tag1$$
Now let:
$$k'=k[\ce{H+}]^x\tag{constant}$$
And then:
$$k' t_\frac 12 = \ln2$$

Is it conceptually correct?

Note:

I have found tons of solutions over the internet all with different answers.

The answer given in my book is x=1 (x is same as is used in eq(1)) and I found the same myself but some sites give answer of the same question as x=2.
We can do this question in two steps, finding the rate law with respect to both A and $\ce{H+}$ individually.
$$\text{Rate}=[A]^y[\ce{H+}]^x$$

The first part is simple, they've given that the half life is independent of $[A]$, therefore it is first order with respect to $\ce{A}$.
Therefore as you've found,
$$\text{Rate}=[A][\ce{H+}]^x$$

Now, onto finding the order with respect to $\ce{H+}$.
From what I've seen on the contradictory answers on the net that OP has mentioned, they state that since the rate is inversely proportional to concentration of $\ce{H+}$, it is second order with respect to $\ce{H+}$ by the relation:
$$t_{1/2}\, \alpha\, [A]^{1-n}$$
where $n$ is the order with respect to A in a reaction.
Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture, i.e.
$$-\frac{\mathrm d[A]}{\mathrm dt}=k[A]^n$$
Here, it is specifically given that it is in a buffer of $\mathrm{pH} = 4$. This means that there is no change in the actual concentration of $\ce{H+}$.
Therefore, what you've done is conceptually sound and is the correct method.
$$k'=k[\ce{H+}]^x = \text{constant}$$
\begin{align}
\frac{t_{1/2}(\mathrm{pH}=5)}{t_{1/2}(\mathrm{pH}=4)} &amp;= \frac{k'(\mathrm{pH}=4)}{k'(\mathrm{pH}=5)} = \frac{k[\ce{H+}]^x(\mathrm{pH}=4)}{k[\ce{H+}]^x(\mathrm{pH}=5)} \\
\implies \frac{100}{10}&amp;=\left(\frac{10^{-4}}{10^{-5}}\right)^x = 10^x \\
\implies x &amp;=1
\end{align}
Therefore, the rate would be:
$$\text{Rate}=k[A][\ce{H+}]$$